# NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.5

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

$$\therefore\space\frac{1}{x(x^2+1)}=\frac{1}{x}+\frac{-x}{x^2+1}\\\therefore\space\int\frac{1}{x(x^2+1)}dx=\\\int\begin{Bmatrix}\frac{1}{x}-\frac{x}{x^2+1}dx\end{Bmatrix}\\=\text{log}|x|-\frac{1}{2}\text{log}(x^2+1)+\text{C}\\\begin{bmatrix}\text{Let x}^2+1=t\Rarr\space 2xdx= dt\\\Rarr x dx=\frac{dt}{2},\\\therefore\space \int\frac{x}{x^2+1}dx=\int\frac{1}{t}\frac{dt}{2}=\frac{1}{2}\text{log} \space t\end{bmatrix}$$