NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.5
Exercise 7.1 Solutions 22 Questions
Exercise 7.2 Solutions 39 Questions
Exercise 7.3 Solutions 24 Questions
Exercise 7.4 Solutions 25 Questions
Exercise 7.5 Solutions 23 Questions
Exercise 7.6 Solutions 24 Questions
Exercise 7.7 Solutions 11 Questions
Exercise 7.8 Solutions 6 Questions
Exercise 7.9 Solutions 22 Questions
Exercise 7.10 Solutions 10 Questions
Exercise 7.11 Solutions 21 Questions
Miscellaneous Exercise on Chapter 7 Solutions 44 Questions
Exercise 7.5
Integrate the rational functions.
$$\textbf{1.}\space\frac{\textbf{x}}{\textbf{(x+1)(x+2)}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{x}{(x+1)(x+2)}dx\\\text{Let,}\space\frac{x}{(x+1)(x+2)}\\\frac{\text{A}}{(x+1)}+\frac{\text{B}}{(x+2)}\\\Rarr\space \frac{x}{(x+1)(x+2)}\\=\frac{A(x+2)+B(x+1)}{(x+1)(x+2)}$$
$$\Rarr\space x=Ax+2A+Bx+B\\\Rarr\space x=x(A+B)+2A+B$$
On equating the coefficients of x and constant term on both sides, we get
A + B = 1 ...(i)
and 2A + B = 0 ...(ii)
On subtracting Eq. (ii) form Eq.(i) we get
$$-\text{A}=1\Rarr\space A=-1$$
Now, on putting the value of A in Eq. (i), we get
$$-1+B=1\\\Rarr\space B=2$$
$$\therefore\space\text{I}=\int\frac{(\normalsize-1)}{x+1}dx+\int\frac{2}{x+2}dx$$
= – log (x + 1) + 2 log (x + 2) + C
= – log (x + 1) + log (x + 2)2 + C
$$=\text{log}\begin{vmatrix}\frac{(x+2)^2}{x+1}\end{vmatrix}+\text{C}\\\bigg[\because\space\text{log b - log a}=log\frac{b}{a}\bigg]$$
$$\textbf{2.}\space\frac{\textbf{1}}{\textbf{x}^\textbf{2}\textbf{-9}}\\\textbf{Sol.}\space\int\frac{1}{x^2-9}dx=\int\frac{1}{x^2-3}dx\\=\int\frac{1}{(x+3)(x-3)}dx\\\text{Let}\space\frac{1}{(x+3)(x-3)}\\=\frac{A}{(x+3)}+\frac{B}{(x-3)}$$
$$\Rarr\space 1=A(x-3)+B(x+3)\\\Rarr\space 1 = x(A+B)+(-3A+3B)$$
On equating the coefficients of x and constant term on both sides, we get
A + B = 0 and – 3A +3B = 1
On solving, we get
$$\text{A}=-\frac{1}{6}\space\text{and B}=\frac{1}{6}\\\therefore\space\int\frac{1}{(x+3)(x-3)}dx\\=\int\frac{(\normalsize-1)}{6(x+3)}dx+\int\frac{1}{6(x-3)}dx\\=-\frac{1}{6}\text{log}|x+3|+\frac{1}{6}\text{log}|x-3|+\text{C}\\=\frac{1}{6}\text{log}\begin{vmatrix}\frac{x-3}{x+3}\end{vmatrix}+\text{C}\\\bigg[\because\space\text{log b - log a}=\text{log}\frac{b}{a}\bigg]$$
$$\textbf{3.}\space\frac{\textbf{3x-1}}{\textbf{(x-1)(x-2)(x-3)}}\\\textbf{Sol.}\space\text{Let}\frac{3x-1}{(x-1)(x-2)(x-3)}\\=\frac{\text{A}}{(x-1)}+\frac{\text{B}}{(x-2)}+\frac{\text{C}}{(x-3)}\\\Rarr\space\frac{3x-1}{(x-1)(x-2)(x-3)}\\=\\\frac{A(x-2)(x-3)+\text{B}(x-1)+\text{C}(x-1)(x-2)}{(x-1)(x-2)(x-3)}$$
$$\Rarr\space 3x – 1=$$
A[x2 – 5x + 6] + B[x2 – 4x + 3] + C[x2 – 3x + 2]
$$\Rarr\space 3x – 1=$$
x2(A + B + C) + x(– 5A – 4B – 3C) + (6A + 3B + 2C)
On equating the coefficients of x2, x and constant term on both sides, we get
A + B + C = 0 ...(i)
– 5A – 4B – 3C = 3 ...(ii)
and 6A + 3B + 2C = – 1 ...(iii)
From eq. (i), we get
A = – (B + C)
On putting the value of A in Eqs. (ii) and (iii), we get
– 5{– (B + C)} – 4B – 3C = 3
$$\Rarr\space 5B+5C-4B-3C=3\\\Rarr\space B + 2C=3\space \text{...(iv)}$$
and 6{– (B + C)} + 3B + 2C = – 1
$$\Rarr\space -6B--6C+3B+2C=-1\\\Rarr\space -3B-4C=-1\space\text{...(v)}$$
On solving equations. (iv) and (v), w
e get
C = 4
On putting the value of C in Eq. (iv), we get
B + 2 × 4 = 3
$$\Rarr\space B=-5$$
Putting the value of B and C in Eq. (i), we get
A + (– 5) + 4 = 0
∴ A = 1, B = – 5, C = 4
$$\text{Now,}\int\frac{3x-1}{(x-1)(x-2)(x-3)}dx\\=\int\bigg(\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}\bigg)dx\\=\int\frac{1}{(x-1)}dx+\int\frac{(-5)}{(x-2)}dx+\\\int\frac{4}{(x-3)}dx$$
= log|x – 1|– 5 log|x – 2| + 4 log|x – 3|+ C
$$\bigg(\because\frac{1}{x}dx= log\space x\bigg)$$
$$\textbf{4.}\space\frac{\textbf{x}}{\textbf{(x-1)(x-2)(x-3)}}\\\textbf{Sol.}\int\frac{x}{(x-1)(x-2)(x-3)}dx\\\text{Let}\space\frac{x}{(x-1)(x-2)(x-3)}\\=\frac{\text{A}}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}\\\Rarr\space\frac{x}{(x-1)(x-2)(x-3)}\\=\\\frac{A(x-2)(x-3)+B(x-1)(x-3)-C(x-1)(x-2)}{(x-1)(x-2)(x-3)}$$
$$\Rarr\space x=A[x^2-2x-3x+6]+\\B[x^2-4x+3]+\text{C}[x^2-x-2x+2]\\\Rarr\space x=x^2(A+B+C)+x(-5A-4B-3C)+\\(6A+3B+2C)$$
On comparing the coefficients of x2, x and constant term on both sides, we get
A + B + C = 0 Þ A = – (B + C) ...(i)
– 5A – 4B – 3C = 1 ...(ii)
and 6A + 3B + 2C = 0 ...(iii)
On putting the value of A in equations (ii) and (iii), we get
– 5[– (B + C)] – 4B – 3C = 1
$$\Rarr\space 5B+5C-4B-3C=1\\\Rarr\space B+2C=1\space\text{...(iv)}\\\text{and 6} \lbrace– (B + C)\rbrace + 3B + 2C = 0\\\Rarr\space -3B-4C=0\space\text{...(v)}$$
Multiplying by 3 in Eq. (iv) and then adding in eq. (v) we get
$$\text{C}=\frac{3}{2}$$
On putting the value of C in Eq. (iv), we get
$$\text{B}+2×\frac{3}{2}=1\\\Rarr\space B=-2.$$
Now, put the values of B and C in Eq. (i), we get
$$A-2+\frac{3}{2}=0\\\Rarr\space A=\frac{4-3}{2}=\frac{1}{2}\\\therefore\space\text{A}=\frac{1}{2}, B=-2\space\text{and C}=\frac{3}{2}\\\text{Now},\space\int\frac{x}{(x-1)(x-2)(x-3)}dx$$
$$=\int\frac{A}{(x-1)}dx+\int\frac{B}{(x-2)}dx+\\\int\frac{C}{(x-3)}dx\\=\frac{1}{2}\int\frac{1}{(x-1)}dx-2\int\frac{1}{(x-2)}dx\\+\frac{3}{2}\int\frac{1}{(x-3)}dx\\=\frac{1}{2}\text{log}|x-1|-2\space\text{log}|x-2|+\\\frac{3}{2}\text{log}|x-3|+\text{C}.$$
$$\textbf{5.}\space\frac{\textbf{2x}}{\textbf{x}^\textbf{2}\textbf{+3x+2}}\\\textbf{Sol.\space}\text{Let}\space\text{I}=\frac{2x}{x^2+3x+2}\\=\frac{2x}{x^2+2x+x+2}\\\frac{2x}{x(x+2)+1(x+2)}=\frac{2x}{(x+2)(x+1)}\\\text{Let}\space\frac{2x}{(x+2)(x+1)}=\frac{A}{(x+2)}+\frac{B}{(x+1)}\\\Rarr\space\frac{2x}{(x+2)(x+1)}\\=\frac{A(x+1)+B(x+2)}{(x+2)(x+1)}$$
$$\Rarr\space 2x=Ax+A+Bx+2B\\\Rarr\space 2x=x(A+B)+ (A+2B).$$
On comparing the coefficients of x and constant term on both sides, we get
A + B = 2 ...(i)
and A + 2B = 0 ...(ii)
On subtracting eq. (ii) form eq. (i), we get
$$-B=2\Rarr\space B=-2$$
On putting the value of B in Eq. (i) we get
$$A-2=2\\\Rarr\space A=4$$
$$\therefore\space\text{I}=\int\frac{A}{(x+2)}dx+\int\frac{B}{(x+1)}dx\\=\int\frac{4}{(x+2)}dx+\int\frac{(\normalsize-2)}{(x+1)}dx$$
= 4 log|x + 2| – 2 log|x + 1| + C
$$\textbf{6.}\space\frac{\textbf{1-x}^\textbf{2}}{\textbf{x(1-2x)}}\textbf{.}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{1-x^2}{x(1-2x)}dx.$$
Here, degree of numerator is equal to degree of denominator, so divide the numerator by denominator.
$$\text{Thus,}\space\frac{1-x^2}{x(1-2x)}\\=\frac{x^2-1}{2x^2-x}\\=\frac{1}{2}+\frac{\frac{1}{2}x-1}{2x^2-x}\\\therefore\space\text{I}=\int\frac{1}{2}dx+\int\frac{\frac{1}{2}x-1}{(2x^2-x)}dx$$
$$\Rarr\space\text{I}=\text{I}_1+\text{I}_2\space\text{...(i)}\\\text{where,}\text{I}_1=\int\frac{1}{2}dx\space\text{and}\space\\\text{I}_2=\int\frac{\frac{1}{2}x-1}{(2x^2-x)}dx\\\text{Now,}\space\text{I}_1=\int\frac{1}{2}dx=\frac{1}{2}x+\text{C}_1\\\text{and}\space\text{I}_2=\int\frac{\frac{1}{2}x+1}{2x^2-x}dx$$
$$\text{Let}\space\frac{\frac{1}{2}-1}{(2x^2-x)}\\=\frac{A}{x}+\frac{B}{(2x-1)}\\\Rarr\\\space\frac{\frac{1}{2}x-1}{(2x^2-x)}=\frac{A(2x-1)+Bx}{x(2x-1)}\\\Rarr\space\frac{1}{2}x-1=2Ax-A+Bx\\\Rarr\space\frac{1}{2}x-1=x(2A+B)-A$$
On comparing the coefficient of x and constant term on both sides, we get
$$2A+B=\frac{1}{2}\space\text{...(i)}$$
and – A = – 1
$$\Rarr\space A=1\space\text{...(ii)}$$
From equation (i),
$$2×1+B=\frac{1}{2}\\\Rarr\space\text{B}=\frac{1}{2}-2=\frac{-3}{2}\\\therefore\space\text{I}_2=\int\bigg[\frac{1}{x}-\frac{3}{2(x-1)}\bigg]dx\\=\int\frac{1}{x}dx-\frac{3}{2}\int\frac{1}{2x-1}dx\\\Rarr\space\text{I}_2=\text{log x}-\frac{3}{2}\frac{log|2x-1|}{2}+\text{C}_2$$
On putting the values of I1 and I2 in Eq (i), we get
$$\text{I}=\frac{1}{2}x+\text{log x}-\frac{3}{4}\text{log}|2x-1|+\text{C}\\\lbrack\text{C}_1+\text{C}_2=\text{C}\rbrack$$
$$\textbf{7.\space}\frac{\textbf{x}}{(\textbf{x}^\textbf{2}\textbf{+1})\textbf{(x-1)}}\\\textbf{Sol.}\space\int\frac{x}{(x^2+1)(x+1)}dx$$
First, we resolve the given integrand into partial fractions.
$$\text{Let}\space\frac{x}{(x^2+1)(x-1)}\\=\frac{A}{x-1}+\frac{BX+C}{x^2+1}\space\text{...(i)}$$
$$\Rarr\space x=A(x^2+1)+(Bx+C)(x-1)\space\text{..(ii)}$$
Substituting x = 1 and 0 in Eq. (ii), we get
1 = A(2) and 0 = A – C
$$\Rarr\space A=\frac{1}{2}\text{and}\space\text{C}=A=\frac{1}{2}$$
On equating the coefficient of x2 on the both sides in equation (ii), we get
$$0=A+B\\\Rarr B=-A=-\frac{1}{2}$$
$$\therefore\space\int\frac{x}{(x^2+1)(x-1)}dx\\=\int\begin{Bmatrix}\frac{\frac{1}{2}}{x-1}+\frac{\bigg(-\frac{1}{2}x+\frac{1}{2}\bigg)}{x^2+1}\end{Bmatrix}dx\\=\frac{1}{2}\int\frac{1}{x-1}dx+\frac{1}{2}\int\frac{(-x+1)}{x^2+1}dx\\=\frac{1}{2}\int\frac{1}{x-1}dx-\frac{1}{4}\int\frac{2x}{x^2+1}dx+\\\frac{1}{2}\int\frac{1}{x^2+1}dx\\=\frac{1}{2}\text{log}|x-1|-\frac{1}{4}\text{I}_1+\frac{1}{2}\text{tan}^{\normalsize-1}x\\+\text{C}_1\space\text{...(iii)}$$
$$\text{where,}\space\text{I}_1=\int\frac{2x}{x^2+1}dx.$$
Let (x2 + 1) = t
$$\Rarr\space 2x dx=dt\\\therefore\space\text{I}_1=\int\frac{dt}{t}=\text{log}|t|+\text{C}_2\\=\text{log}|x^2+1|+\text{C}_2$$
On putting these values in eq. (iii), we get
$$\int\frac{x}{(x^2+1)(x-1)}dx\\=\frac{1}{2}\text{log}|x-1|-\frac{1}{4}\text{log}|x^2+1|\\+\frac{1}{2}\text{tan}^{\normalsize-1}x+\text{C}$$
[C = C1 + C2]
$$\textbf{8.}\space\frac{\textbf{x}}{\textbf{(x-1)}^\textbf{2}\textbf{(x+2)}}\\\textbf{Sol.}\space\text{Let}\space\frac{x}{(x-1)^2(x+2)}\\=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{\text{C}}{x+2}\\\text{...(i)}$$
$$\Rarr\space x=A(x-1)(x+2)+\text{B}(x+2)+\\\text{C}(x-1)^2\text{...(ii)}\\\text{On substituting x = 1, – 2 in eq. (ii), we get}\\\text{1 = B(1 + 2) and – 2 = C(– 2 – 1)}^2$$
$$\Rarr\space\text{B}=\frac{1}{3}\space\text{and}\text{C}=\frac{-2}{9}$$
On equating the coefficient of x2 on both sides in equation (ii), we get
0 = A + C
$$\Rarr\space\text{A}=-\text{C}=\frac{2}{9}\\\therefore\space\int\frac{x}{(x-1)^2(x+2)}dx\\=\int\begin{Bmatrix}\frac{\frac{2}{9}}{x-1}+\frac{\frac{1}{3}}{(x-1)^2}+\frac{\bigg(-\frac{2}{9}\bigg)}{x+2}\end{Bmatrix}dx\\\text{(from Eq. (i))}\\\Rarr\space\int\frac{x}{(x-1)^2(x+2)}dx\\=\frac{2}{9}\int\frac{1}{x-1}dx+\frac{1}{3}\int\frac{1}{(x-1)^2}dx-\\\frac{2}{9}\int\frac{1}{x+2}dx$$
$$=\frac{2}{9}\text{log}|x-1|+\frac{1}{3}\frac{(x-1)^{-2+1}}{(-2+1)}-\\\frac{2}{9}\text{log}|x+2|+\text{C}\\=\frac{2}{9}\text{log}\begin{vmatrix}\frac{x-1}{x+2}\end{vmatrix}-\frac{1}{3}\bigg(\frac{x}{x-1}\bigg)+\text{C}\\\bigg[\because\space\text{log b - log a = log}\bigg(\frac{b}{a}\bigg)\bigg]$$
$$\textbf{9.}\space\frac{\textbf{3x+5}}{\textbf{x}^\textbf{3}\textbf{-x}^\textbf{2}\textbf{-x+1}}\\\textbf{Sol.}\space\int\frac{3x+5}{x^3-x^2-x+1}dx\\=\int\frac{3x+5}{(x^2-1)(x-1)}dx\\=\int\frac{3x+5}{(x-1)(x+1)(x-1)}dx\\=\int\frac{3x+5}{(x-1)^2(x+1)}dx\\\text{Let}\space\frac{3x+5}{(x-1)^2(x+1)}\\=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}$$
$$\Rarr\space\frac{3x+5}{(x-1)^2(x+1)}\\=\frac{A(x-1)(x+1)+B(x+1)+C(x-1)^2}{(x-1)^2(x+1)}\\\Rarr\space 3x+5\\=A(x^2-1)+Bx+B+C(x^2+1-2x)\\\Rarr\space 3x+5\\=x^2(A+C)+x(B-2C)+(-A+B+C)$$
On comparing the coefficients of x2, x and constant term on both sides, we get
A + C = 0
$$\Rarr\space A=-C\space\text{...(i)}$$
B – 2C = 3 ...(ii)
and – A + B + C = 5 ...(iii)
On putting the value of A from eq. (i) in eq. (iii), we get
– (– C) + B + C = 5
$$\Rarr\space B+2C=5\space\text{...(iv)}$$
On adding eq. (ii) and eq. (iv), we get
2B = 8
$$\Rarr\space B=4$$
On putting the value of B in eq. (ii), we get
4 – 2C = 3
$$\Rarr\space 1=2C\Rarr\space \text{C}=\frac{1}{2}\\\text{Also,}\space \text{A=-C}=-\frac{1}{2}\\\therefore\space\int\frac{3x+5}{(x-1)^2(x+1)}dx\\=\int\bigg(\frac{-\frac{1}{2}}{x-1}+\frac{4}{(x-1)^2}+\frac{\frac{1}{2}}{x+1}\bigg)dx$$
$$=-\frac{1}{2}\int\frac{1}{x-1}dx+4\int\frac{1}{(x-1)^2}dx+\\\frac{1}{2}\int\frac{1}{x+1}dx\\=\frac{1}{2}\text{log}|x-1|+4\frac{(x-1)^{-2+1}}{(-2+1)}+\\\frac{1}{2}\text{log}|x+1|+\text{C}\\=\frac{1}{2}\text{log}\begin{vmatrix}\frac{x+1}{x-1}\end{vmatrix}-\frac{4}{x-1}+\text{C}\\\bigg[\because\space\text{log b - log a}=log\bigg(\frac{b}{a}\bigg)\bigg]$$
$$\textbf{10.}\space\frac{\textbf{2x+3}}{\textbf{(x}^\textbf{2}\textbf{-1})\textbf{(2x+3)}}\\\textbf{Sol.}\space\int\frac{2x-3}{(x^2-1)(2x+3)}dx\\=\int\frac{2x-3}{(x-1)(x+1)(2x+3)}dx\\\text{Let}\space\frac{2x-3}{(x-1)(x+1)(2x+3)}\\=\frac{A}{(x-1)}+\frac{B}{(x+1)}+\frac{\text{C}}{(2x+3)}\\\Rarr\space\frac{2x-3}{(x-1)(x+1)(2x+3)}\\=\\\frac{A(2x+3)(x+1)+B(x-1)(2x+3)+\text{C}(x-1)(x+1)}{(x-1)(x+1)(2x+3)}$$
$$\Rarr\space\text{2x-3}=A(2x^2+3x+2x+3)+\\\text{B}(2x^2-2x+3x-3)+\text{C}(x^2-1)\\\Rarr\space 2x-3\\=x^2(2A+2B+C)+x(5A+B)+\\(3A-3B-C)$$
On comparing the coefficients of x2, x and constant term on both sides, we get
2A + 2B + C = 0 ...(i)
5A + B = 2
$$\Rarr\space B=2-5A\space\text{...(ii)}$$
and 3A – 3B – C = – 3 ...(iii)
On putting the value of B in Eqs. (i) and (iii), we get
2A + 2(2 – 5A) + C = 0 Þ 2A + 4 – 10A + C = 0
$$\Rarr\space -8A+C=-4\space\text{...(iv)}\\\text{and}\space\text{3A-3}(2-5A)-C=-3\\\Rarr\space 3A-6+15 A-C=-3\\\Rarr\space 18 A-C=3\space\text{...(v)}$$
On adding equations (iv) and (v), we get
$$10A=-1\\\Rarr\space A=\frac{-1}{10}$$
On putting the value of A in equations (ii), we get
$$2\bigg(-\frac{1}{10}\bigg)+2\bigg(\frac{5}{2}\bigg)+\text{C}=0\\\Rarr-\frac{1}{5}+5+\text{C}=0\\\Rarr\space \text{C}=-\frac{24}{5}\\\therefore\space\text{A}=-\frac{1}{10},\text{B}=\frac{5}{2}\space\text{ans C}=-\frac{24}{5}\\\therefore\space\int\frac{2x-3}{(x^2-1)(2x+3)}dx\\=\int\frac{(\normalsize-1)}{10(x-1)}dx+\frac{5}{2}\int\frac{1}{x+1}dx\\-\frac{24}{5}\int\frac{1}{2x+3}dx$$
$$=\frac{5}{2}\text{log}|x+1|-\frac{1}{10}\text{log}|x-1|-\\\frac{12}{5}\text{log}|2x+3|+\text{C}$$
$$\textbf{11.}\space\frac{\textbf{5x}}{\textbf{(x+1)}\textbf{(x}^\textbf{2}\textbf{-4)}}\\\textbf{Sol.}\space\int\frac{5x}{(x+1)(x^2-4)}dx\\=\int\frac{5x}{(x+1)(x+2)(x-2)}dx\\\text{Let}\space \frac{5x}{(x+1)(x-2)(x+2)}\\=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)}\\=\\\frac{A(x+2)(x-2)+B(x+1)(x-2)+\text{C}(x+1)(x+2)}{(x+1)(x+2)(x-2)}$$
$$\Rarr\space 5x=A(x^2-4)+B(x^2+x-2x-2)\\+\text{C}(x^2+x+2x+2)\\\Rarr\space 5x=x^2(A+B+C)+x(-B+3C)+\\(-4A-2B+2C)$$
On comparing the coefficients of x2, x and constant term on both sides, we get
A + B + C = 0 ...(ii)
– B + 3C = 5 ...(iii)
and – 4A – 2B + 2C = 0 ...(iv)
Multiply by 4 in eq. (ii) and then adding with eq. (iv), we get
2 (B + 3C) = 0
$$\Rarr\space B+3C=0\space\text{...(v)}$$
On adding equations (iii) and (v), we get
$$\text{C}=\frac{5}{6}$$
On putting the value of C in equations (v), we get
$$\text{B}=-\frac{5}{2}$$
On putting the values of B and C in Eq. (ii), we get
$$\text{A}=\frac{5}{3}\\\therefore\space\int\frac{5x}{(x+1)(x-2)(x+2)}dx\\=\int\frac{5}{3(x+1)}dx-\int\frac{5}{2(x+2)}dx+\\\int\frac{5}{6(x-2)}dx\\=\frac{5}{3}\int\frac{1}{(x+1)}dx-\frac{5}{2}\int\frac{1}{(x+2)}dx+\\\frac{5}{6}\int\frac{1}{(x-2)}dx\\=\frac{5}{3}\text{log}|x+1|-\frac{5}{2}\text{log}|x+2|+\\\frac{5}{6}\text{log}|x-2|+\text{C}$$
$$\textbf{12.}\space\frac{\textbf{x}^\textbf{3}\textbf{+x+1}}{\textbf{x}^\textbf{2}\textbf{-1}}\\\textbf{Sol.}\space\int\frac{x^3+x+1}{x^2-1}dx\\\int x dx + \int\frac{2x+1}{x^2+1}dx\\\bigg[\because\space \frac{x^3+x+1}{x^2-1}=x+\frac{2x+1}{x^2-1}\bigg]\\=\int x dx + \int\frac{2x+1}{(x+1)(x-1)}dx\\\text{Let}\space\frac{2x+1}{(x+1)(x-1)}=\\\frac{A}{(x+1)}+\frac{B}{(x-1)}$$
$$\Rarr\space \frac{2x+1}{(x+1)(x-1)}=\\\frac{A(x-1)+B(x+1)}{(x+1)(x-1)}\\\Rarr\space 2x+1 = x(A+B)-A+B$$
On comparing the coefficients of x and constant term on both sides, we get
A + B = 2 and – A + B = 1
On adding above equations, we get
$$2B=3\Rarr\text{B}=\frac{3}{2}\\\text{and then A}=\frac{1}{2}\\\therefore\space\int\frac{x^2+x+1}{x^2-1}dx=\\\int x dx+\int\frac{A}{(x+1)}dx+\int\frac{B}{(x-1)}dx\\=\int xdx + \frac{1}{2}\int\frac{1}{(x+1)}dx+\\\frac{3}{2}\int\frac{1}{(x-1)}dx\\=\frac{x^2}{2}+\frac{1}{2}\text{log}|x+1|+\\\frac{3}{2}\text{log}|x-1|+\text{C}$$
$$\textbf{13.}\space\frac{\textbf{2}}{\textbf{(1-x)(1+x}^\textbf{2}\textbf{)}}\\\textbf{Sol.}\space\text{Let}\space\frac{2}{(1-x)(1-x^2)}=\\\frac{A}{1-x}+\frac{Bx+C}{1+x^2}\\\therefore\space \frac{2}{(1-x)(1+x^2)}=\\\frac{A(1+x^2)+(Bx+C)(1-x)}{(1-x)(1-x^2)}$$
$$\Rarr 2 = A + Ax^2 + Bx + C – Bx^2 – Cx\\\Rarr 2 = x^2(A – B) + x(B – C) + (A + C)$$
On comparing the coefficients of x2, x and constant term on both sides, we get
A – B = 0
$$\Rarr\space A=B\space \text{...(i)}$$
B – C = 0
$$\Rarr\space B=C\space\text{...(ii)}\\\text{and}\space A + C=2\space\text{...(iii)}$$
From Eqs. (i) and (ii), we get A = C put this value in Eq. (iii), we get
$$\text{2A=2}\Rarr\space A=1$$
Put the value of A in Eqs. (i) and (iii), we get
B = 1 and C = 1
$$\therefore\space\int\frac{2}{(1-x)(1+x^2)}dx\\=\int\begin{Bmatrix}\frac{1}{1-x} + \frac{x+1}{x^2+1}\end{Bmatrix}dx\\\int\frac{1}{1-x}dx+\frac{1}{2}\int\frac{2x}{x^2+1}dx\\\int\frac{1}{x^2+1}dx\\=-\text{log}|1-x|+\frac{1}{2}\text{log}(1+x^2)+\\\text{tan}^{\normalsize-1}x+\text{C}$$
[Let x2 + 1 = t
$$\Rarr\space 2x dx=dt\\\therefore\space \int\frac{2x}{x^2+1}dx=\int\frac{1}{t}dt\text{log t}]$$
$$\textbf{14.}\space\frac{\textbf{3x-1}}{\textbf{(x+2)}^\textbf{2}}\\\textbf{Sol.}\space\text{Let}\space \frac{3x-1}{(x+2)^2}=\\\frac{A}{(x+2)}+\frac{B}{(x+2)^2}$$
$$\Rarr\space 3x-1=A(x+2)+B$$
On equating the coefficients of x and constant term on both sides, we get
A = 3 and 2A + B = – 1
∴ 2(3) + B = – 1
$$\Rarr\space B=-7\\\therefore\space\frac{3x-1}{(x+2)^2}=\frac{3}{(x+2)}-\frac{7}{(x+2)^2}\\\therefore\space \int\frac{3x-1}{(x+2)^2}=\\3\int\frac{1}{(x+2)}dx-7\int\frac{1}{(x+2)^2}dx\\= 3\text{log}|x+2|-7\bigg(\frac{-1}{x+2}\bigg)+\text{C}\\= 3\text{log}|x+2|+\frac{7}{x+2}+\text{C}$$
$$\textbf{15.}\space\frac{\textbf{1}}{\textbf{x}^\textbf{4}\textbf{-1}}\\\textbf{Sol.}\space \int\frac{1}{x^4-1}dx=\int\frac{1}{(x^2-1)(x^2+1)}dx$$
[∵ a2 – b2 = (a + b) (a – b)]
$$\text{Let}\space\frac{1}{(x^2-1)(x^2+1)}=\\\frac{\text{Ax+B}}{(x^2+1)}+\frac{\text{Cx+D}}{(x^2-1)}\\\Rarr\space \frac{1}{(x^2+1)(x^2-1)}\\=\\\frac{(Ax+B)(x^2-1) + (Cx+D)(x^2+1)}{(x^2+1)(x^2-1)}$$
$$\Rarr\space 1=Ax^3+Bx^2-Ax-B+Cx^3+\\x^2D+Cx+D\\\Rarr\\ 1=x^3(A+C)+x^2(B+D)+\\x(-A+C) + (-B+D)$$
On comparing the coefficients of x3, x2, x and constant term on both sides, we get
A + C = 0 ...(i)
B + D = 0 ...(ii)
– A + C = 0 ...(iii)
and – B + D = 1 ...(iv)
On adding equations (i) and (iii), w e get
2C = 0
$$\Rarr\space C=0\Rarr\space C=0,\\\text{then A = 0}[\text{from Eq. (i)}]$$
On adding equations (ii) and (iv), we get
$$\text{2D = 1}\\\Rarr\space D=\frac{1}{2}\\\text{On putting the value of D in Eq. (iv), we get}\\-\text{B}+\frac{1}{2}\\\Rarr\space\text{B}=-\frac{1}{2}\\\therefore\space \text{A = 0,}\space \text{B}=-\frac{1}{2},\text{C=0}\\\text{and D}=\frac{1}{2}$$
$$\therefore\space\int\frac{1}{(x^4-1)}dx=\int\frac{Ax+B}{(x^2+1)}dx\\+\int\frac{\text{Cx+D}}{(x^2-1)}dx\\=-\frac{1}{2}\int\frac{1}{x^2+1}dx+\frac{1}{2}\int\frac{1}{x^2-1}dx\\=-\frac{1}{2}\text{tan}^{\normalsize-1}x+\frac{1}{2}.\frac{1}{2}\text{log}\begin{vmatrix}\frac{x-1}{x+1}\end{vmatrix}+\text{C}\\=-\frac{1}{2}\text{tan}^{\normalsize-1}x+\frac{1}{4}\text{log}\begin{vmatrix}\frac{x-1}{x+1}\end{vmatrix}+\text{C}$$
$$\textbf{16.}\space\frac{\textbf{1}}{\textbf{x(x}^\textbf{n}\textbf{+1)}}\\\textbf{Sol.}\space\text{Let\space}\text{I}=\int\frac{1}{x(x^n+1)}dx\\=\int\frac{x^{n-1}}{x^n(x^n+1)}dx\\\text{Put x}^n=t\\\Rarr nx^{n-1}dx=dt\\\Rarr\space x^{n-1}dx=\frac{1}{n}dt\\\therefore\space\text{I}=\int\frac{x^{n-1}}{x^n(x^n+1)}dx\\=\frac{1}{n}\int\frac{1}{t(t+1)}dt\space\text{...(i)}$$
$$\text{Now,}\space\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{(t+1)}\\\Rarr\space 1=A(1+t)+Bt\space\text{...(ii)}$$
On substituting t = 0, – 1 in eq. (ii), we get
A = 1 and B = – 1
$$\therefore\space\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{(t+1)}\\\therefore\space\text{I}=\frac{1}{n}\int\bigg(\frac{1}{t}-\frac{1}{(t+1)}\bigg)dt\\\lbrack\text{from eq.(i)}\rbrack\\=\frac{1}{n}[\text{log} |t|-\text{log}|t+1|]+\text{C}\\=\frac{1}{n}[\text{log}|x^n|-\text{log}|x^n+1|]+\text{C}\\\lbrack\text{put t = x}^n\rbrack\\=\frac{1}{n}\text{log}\begin{vmatrix}\frac{x^n}{x^n+1}\end{vmatrix}+\text{C}$$
$$\textbf{17.}\space \frac{\textbf{cos x}}{\textbf{(1 - sin x)(2 - sin x)}}$$
[Hint : Put sin x = t]
$$\textbf{Sol.}\space\text{Let}\\\space\text{I}=\int\frac{\text{cos x}}{(1 - sin x)(2 - sin x)}dx\\\text{Put}\space\text{sin x=t}\\\Rarr\text{cos x}=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{\text{cos x}}\\\therefore\space \text{I}=\int\frac{\text{cos x}}{(1-t)(2-t)}\frac{dt}{\text{cos x}}\\=\int\frac{1}{(1-t)(2-t)}dt\\=\int\bigg[\frac{A}{1-t} + \frac{B}{2-t}\bigg]dt\\\text{...(i)}$$
$$\therefore\space\frac{1}{(1-t)(2-t)}=\frac{A(2-t)+B(1-t)}{(1-t)(2-t)}$$
$$\Rarr\space 1=2A-tA+B-Bt\\\Rarr\space 1=1(2A+B) +t(-A-B)$$
On comparing the coefficients of t and constant term on both sides, we get
2A + B = 1 and – A – B = 0
On adding above equations, we get
A = 1 and then B = – 1
$$\therefore\space\text{I}=\int\bigg(\frac{1}{1-t}-\frac{1}{2-t}\bigg)dt\\\lbrack\text{from eq. (i)}\rbrack\\=\int\frac{1}{(1-t)}dt-\int\frac{1}{(2-t)}dt\\=\frac{\text{log}|1-t|}{(-1)}-\frac{\text{log}|2-t|}{(-1)}+\text{C}\\=\text{log}\begin{vmatrix}\frac{2-t}{1-t}\end{vmatrix}+\text{C}\\=\text{log}\begin{vmatrix}\frac{2- \text{sin x} }{1-\text{sin x}}\end{vmatrix}+\text{C}$$
(Put t = sin x)
$$\textbf{18.}\space\frac{\textbf{(x}^\textbf{2}\textbf{+1)(x}^\textbf{2}\textbf{+2)}}{\textbf{(x}^\textbf{2}\textbf{+3)}\textbf{(x}^\textbf{2}\textbf{+4)}}$$
Sol. Here, the degree of numerator and denominator are 4. So, we convert it into simple form by putting x2 = t.
$$\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=\\\frac{(t+1)(t+2)}{(t+3)(t+4)}=\frac{t^2+3t+2}{t^2+7t+12}$$
Since, degree of numerator and denominator is same, so it can be written as
$$=1-\frac{4t+10}{t^2+7t+12}\\\text{Now},\int\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx\\=\int 1dx-\int\frac{(4t+10)}{t^2+7t+12}dx\\=\int 1 dx-\int\frac{4t-10}{(t+3)(t+4)}dx\\\space\text{...(i)}\\\text{Let}\space\frac{4t+10}{(t+3)(t+4)}=\\\frac{A}{(t+4)}+\frac{B}{(t+3)}\\\Rarr\space \frac{4t+10}{(t+3)(t+4)}=\frac{A(t+3)+B(t+4)}{(t+4)(t+3)}$$
$$\Rarr\space 4t+10=At+3A+Bt+4B\\\Rarr\space 4t+10=t(A+B)+(3A+4B)$$
On comparing the coefficients of t and constant term on both sides, we get
A + B = 4
$$\Rarr\space 3A + 3B=12\space\text{...(ii)}$$
and 3A + 4B = 10 ...(iii)
On subtracting Eq. (iii) from Eq. (ii), w
e get
– B = 2 and 3A + 4B = 10 ...(iii)
On subtracting Eq. (iii) from Eq. (ii), we get
– B = 2
$$\Rarr\space\text{B = -2}\space\text{and A=6}$$
$$\therefore\space\frac{4t+10}{(t+4)(t+3)}=\frac{6}{t+4}-\frac{2}{t+3}$$
On putting this value in eq. (i), we get
$$\int\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx=\\\int 1 dx-\int\begin{bmatrix}\frac{6}{t+4} - \frac{2}{t+3}\end{bmatrix}dx\\=\int 1dx-\int\bigg(\frac{6}{(x^2+4)}-\frac{3}{(x^2+3)}\bigg)dx\\\lbrack\text{Put} \space t=x^2\rbrack\\=\int\text{1 dx}-\int\bigg(\frac{6}{x^2+2^2}-\frac{2}{x^2 + (\sqrt{3})^2} \bigg)dx\\= x-6\bigg(\frac{1}{2}\text{tan}^{\normalsize-1}\frac{x}{2}\bigg)+2\bigg(\frac{1}{\sqrt{3}}\text{tan}^{\normalsize-1}\frac{x}{\sqrt{3}}\bigg)\\+\text{C}\\\bigg(\because\space \int\frac{1}{a^2+x^2}dx=\frac{1}{a}\text{tan}^{\normalsize-1}\frac{x}{a}\bigg)$$
$$= x-3\space\text{tan}^{\normalsize-1}\frac{x}{2}+\frac{2}{\sqrt{3}}\text{tan}^{\normalsize-1}\frac{x}{\sqrt{3}}+\text{C}$$
$$\textbf{19.}\space\frac{\textbf{2x}}{(\textbf{x}^\textbf{2}\textbf{+1)(x}^\textbf{2}\textbf{+3)}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{2x}{(x^2+1)(x^2+3)}dx\\\text{Put x}^2=t\\\Rarr 2x=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{2x}\\\therefore\space\text{I}=\int\frac{2x}{(t+1)(t+3)}\frac{dt}{2x}\\=\int\frac{1}{(t+1)(t+3)}dt$$
$$\text{Let}\space \frac{1}{(t+1)(t+3)}=\\\frac{A}{t+1}+\frac{B}{t+3}\\\Rarr\space \frac{1}{(t+1)(t+3)}=\frac{At+3A+Bt+B}{(t+1)(t+3)}$$
$$\Rarr 1 = (A + B)t + (3A + B)$$
On comparing the coefficients of t and constant terms on both sides, we get
A + B = 0 ...(i)
and 3A + B = 1 ...(ii)
On subtracting eq. (ii) from eq. (i), we get
$$2A=1\\\Rarr\space \text{A}=\frac{1}{2}$$
On putting the value of A in eq. (i), we get
$$\text{B}=-\frac{1}{2}\\\therefore\space \text{I}=\frac{1}{2}\int\frac{1}{t+1}dt-\frac{1}{2}\int\frac{1}{t+3}dt\\=\frac{1}{2}\text{log}|t+1|-\frac{1}{2}\text{log}|t+3|+\text{C}\\=\frac{1}{2}\text{log}\begin{vmatrix}\frac{t+1}{t+3}\end{vmatrix}+\text{C}\\=\frac{1}{2}\text{log}\begin{vmatrix}\frac{x^2+1}{x^2+3}\end{vmatrix}+\text{C}\\\lbrack\text{Put t=x}^2\rbrack$$
$$\textbf{20.}\space\frac{\textbf{1}}{\textbf{x(x}^\textbf{4}\textbf{-1)}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{1}{x(x^4-1)}dx\\=\int\frac{x^3}{x^4(x^4-1)}dx\\=\frac{1}{4}\int\frac{4x^3}{x^4(x^4-1)}dx\\\text{Put x}^4=t\\\Rarr\space 4x^3=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{4x^3}\\\therefore\space \text{I}=\frac{1}{4}\int\frac{4x^3}{t(t-1)}\frac{dt}{4x^3}$$
$$\frac{1}{4}\int\frac{dt}{t(t-1)}$$
$$\text{Let}\space\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$$
Þ 1 = A(t – 1) + Bt ...(i)
On substituting t = 0 and 1 in eq. (i), we get
A = – 1 and B = 1
$$\Rarr\space 1=A(t-1)+Bt\space\text{...(i)}$$
On substituting t = 0 and 1 in eq. (i), we get
A = – 1 and B = 1
$$\therefore\space \frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}\\\therefore\space\text{I}=\frac{1}{4}\int\bigg(-\frac{1}{t}+\frac{1}{t-1}\bigg)dt\\=\frac{1}{4}\lbrace- log |t| + \text{log} |t-1|+\text{C}\rbrace\\=\frac{1}{4}\text{log}\begin{vmatrix}\frac{t-1}{t}\end{vmatrix}+\text{C}\\=\frac{1}{4}\text{log}\begin{vmatrix}\frac{x^4-1}{x^4}\end{vmatrix}+\text{C}$$
(Put t = x4)
$$\textbf{21.}\space\frac{\textbf{1}}{\textbf{e}^\textbf{x}\textbf{-1}}\qquad \lbrack\textbf{Hint :}\space \text{Put} e^x=t\rbrack\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{1}{e^x-1}dx$$
On multiplying numerator and denominator by e–x, we get
$$\text{I}=\int\frac{e^{\normalsize-x}}{1-e^{\normalsize-x}}dx\\\text{Put}\space 1-e^{-x}=t\\\Rarr\space -e^{-x}(-1)=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{e^{\normalsize-x}}\\\Rarr\space e^{-x}dx=dt\\\therefore\space \text{I}=\int\frac{dt}{t}=\text{log}|t|+\text{C}\\=\text{log}|1-e^{-x}|+\text{C}\\=\text{log}\begin{vmatrix}\frac{e^x-1}{e^x}\end{vmatrix}+\text{C}$$
Choose the correct answer in each of the question.
$$\textbf{22.}\int\space\frac{\textbf{x}}{\textbf{(x-1)(x-2)}}\textbf{dx}\space\textbf{equals}\\\textbf{(a)}\space\textbf{log}\begin{vmatrix}\frac{\textbf{(x-1)}^\textbf{2}}{\textbf{x-2}}\end{vmatrix}+\textbf{C}\\\textbf{(b)}\space\textbf{log}\begin{vmatrix}\frac{\textbf{(x-2)}^\textbf{2}}{\textbf{x-1}}\end{vmatrix}+\textbf{C}\\\textbf{(c)}\space\textbf{log}\begin{vmatrix}\bigg(\frac{\textbf{x-1}}{\textbf{x-2}}\bigg)\end{vmatrix}\\\textbf{(d)}\space\textbf{log}\textbf{|(x-1)(x-2)|}+\textbf{C}\\\textbf{Sol.}\space (b)\space\text{log}\begin{vmatrix}\frac{(x-2)^2}{x-1}\end{vmatrix}+\text{C}\\\text{Let}\space\frac{x}{(x-1)(x-2)}=\\\frac{A}{(x-1)}+\frac{B}{(x-2)}$$
$$\Rarr\space x = A(x-2)+B(x-1)\space\text{...(i)}$$
On substituting x = 1 and 2 in Eq. (i), we get
A = – 1 and B = 2
$$\therefore\space\frac{x}{(x-1)(x-2)}\\=-\frac{A}{(x-1)}+\frac{B}{(x-1)}\\\therefore\space\int\frac{x}{(x-1)(x-2)}dx=\\\int\frac{(-1)}{x-1}dx+\int\frac{2}{x-2}dx$$
= – log|x – 1| + 2 log|x – 2| + C
= – log|x – 1| + log|x – 2|2 + C
$$=\text{log}\begin{vmatrix}\frac{(x-2)^2}{x-1}\end{vmatrix}+\text{C}\\\bigg[\because\space \text{log b - log a}=log\frac{b}{a}\bigg]$$
$$\textbf{23.}\space\int\frac{\textbf{dx}}{\textbf{x(x}^\textbf{2}\textbf{+1)}}\space\textbf{equals}\textbf{:}\\\textbf{(a)\space log}|\textbf{x}|-\frac{\textbf{1}}{\textbf{2}}\textbf{log}\textbf{(}\textbf{x}^\textbf{2}\textbf{+1}\textbf{)}\\\textbf{(b)}\space \textbf{log}|\textbf{x}|+\frac{\textbf{1}}{\textbf{2}}\space\textbf{log}(\textbf{x}^\textbf{2}\textbf{+1})\textbf{+C}\\\textbf{(c)\space}\textbf{-log}\textbf{|x|}\textbf{+}\frac{\textbf{1}}{\textbf{2}}\textbf{log}(\textbf{x}^\textbf{2}\textbf{+1})\textbf{+}\textbf{C}\\\textbf{(d)}\space\frac{\textbf{1}}{\textbf{2}}\textbf{log}|\textbf{x}|\textbf{+ log}(\textbf{x}^\textbf{2}\textbf{+1})\textbf{+ C}$$
$$\textbf{Sol.}\space\text{(a) log}|x|-\frac{1}{2}\text{log}(x^2+1)+\text{C}\\\text{Let}\space\frac{1}{x(x^2+1)}\\=\frac{A}{x}+\frac{Bx+C}{x^2+1}$$
$$\Rarr\space 1 = A(x^2+1)+(Bx+C)x$$
On equating the coefficients of x2, x and constant term on both sides, we get
A + B = 0, C = 0 and A = 1
On solving these equations, we get
A = 1, B = – 1 and C = 0
$$\therefore\space\frac{1}{x(x^2+1)}=\frac{1}{x}+\frac{-x}{x^2+1}\\\therefore\space\int\frac{1}{x(x^2+1)}dx=\\\int\begin{Bmatrix}\frac{1}{x}-\frac{x}{x^2+1}dx\end{Bmatrix}\\=\text{log}|x|-\frac{1}{2}\text{log}(x^2+1)+\text{C}\\\begin{bmatrix}\text{Let x}^2+1=t\Rarr\space 2xdx= dt\\\Rarr x dx=\frac{dt}{2},\\\therefore\space \int\frac{x}{x^2+1}dx=\int\frac{1}{t}\frac{dt}{2}=\frac{1}{2}\text{log} \space t\end{bmatrix}$$
