EXERCISE 7.5(Integration)

Integrate the rational functions in Exercises 1 to 21.(Ex 7.5 integration ncert maths solution class 12)

Question 1: $\displaystyle\int \dfrac{x}{(x+1)(x+2)}dx$

Solution: $\dfrac{x}{(x+1)(x+2)}= \dfrac{A}{x+1}+\dfrac{B}{x+2}$

$\Rightarrow \dfrac{x}{(x+1)(x+2)}=\dfrac{A(x+2)+B(x+1)}{(x+1)(x+2)}$

$\Rightarrow x = A(x+2)+B(x+1)--(i)$

Let $x = -1$ putting in (i)

$\Rightarrow -1= A(-1+2)+B(-1+1)$

$\Rightarrow -1=A$

Again putting $x = -2$

$\Rightarrow -2 = A(-2+2)+B(-2+1)$

$\Rightarrow -2 = 0-B$

$\Rightarrow B = 2$

Now, Integrating

Let $I=\displaystyle \int\dfrac{x}{(x+1)(x+2)}dx= \displaystyle\int\dfrac{-1}{x+1}dx+ \displaystyle\int\dfrac{2}{x+2}dx$

$= -\log|x+1|+2\log|x+2|+C$

$=\log(x+2)^2-\log(x+1)+C$

$I = \log\dfrac{(x+2)^2}{(x+1)}+C$

Question 2: $\displaystyle\int \dfrac{1}{(x^2-9)}dx$

Solution: $\dfrac{1}{(x^2-9)}=\dfrac{1}{(x-3)(x+3)}$

$\dfrac{1}{(x-3)(x+3)}=\dfrac{A}{(x-3)}+\dfrac{B}{(x+3)}$

$\Rightarrow \dfrac{1}{(x-3)(x+3)}=\dfrac{A(x+3)+B(x-3)}{(x-3)(x+3)}$

$\Rightarrow 1= A(x+3)+B(x-3)--(i)$

Let $x=3$

$1 = A(3+3)+B(3-3)$

$\Rightarrow 1 = 6A+0$

$\Rightarrow A = \dfrac{1}{6}$

Let $x=-3$

$1 = A(-3+3)+B(-3-3)$

$\Rightarrow 1 = 0-6B$

$\Rightarrow B = -\dfrac{1}{6}$

Now

$\dfrac{1}{(x-3)(x+3)}=\dfrac{1/6}{(x-3)}+\dfrac{-1/6}{(x+3)}$

Integrating

$\displaystyle\int\dfrac{1}{(x-3)(x+3)}dx=\displaystyle\int\dfrac{1/6}{(x-3)}dx+\displaystyle\int\dfrac{-1/6}{(x+3)}dx$

$=\dfrac{1}{6}\log|x-3|-\dfrac{1}{6}\log|x+3+C$

$=\dfrac{1}{6}\log\left|\dfrac{x-3}{x+3}\right|+C$

Question 3: $\displaystyle\int \dfrac{3x-1}{(x-1)(x-2)(x-3)}dx$

Solution: $\dfrac{3x-1}{(x-1)(x-2)(x-3)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-2)}+\dfrac{C}{(x-3)}--(i)$

$\Rightarrow \dfrac{3x-1}{(x-1)(x-2)(x-3)}=\dfrac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}$

$\Rightarrow (3x-1)= A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$

Let $x = 1$

$3\times 1-1 = A(1-2)(1-3)+0+0$

$\Rightarrow 2=A\times -1\times -2$

$\Rightarrow A = 1$

Again Let $x= 2$

$3\times 2-1 = 0+B(2-1)(2-3)+0$

$\Rightarrow 5=B\times 1\times -1$

$\Rightarrow B = -5$

Again let $x = 3$

$3\times 3-1 =0+ C(3-1)(3-2)+0$

$\Rightarrow 8=C\times 2\times 1$

$\Rightarrow C = 4$

Putting the value of A, B and C in eq (i)

$\dfrac{3x-1}{(x-1)(x-2)(x-3)}=\dfrac{1}{(x-1)}+\dfrac{-5}{(x-2)}+\dfrac{4}{(x-3)}$

Integrating

$\displaystyle\int\dfrac{3x-1}{(x-1)(x-2)(x-3)}dx=\displaystyle\int\dfrac{1}{(x-1)}dx+\displaystyle\int\dfrac{-5}{(x-2)}dx+\displaystyle\int\dfrac{4}{(x-3)}dx$

$=\log|x-1|-5\log|x-2|+4\log|x-3+C$

Question 4: $\displaystyle\int \dfrac{x}{(x-1)(x-2)(x-3)}dx$

Solution: $\dfrac{x}{(x-1)(x-2)(x-3)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-2)}+\dfrac{C}{(x-3)}--(i)$

$\Rightarrow \dfrac{x}{(x-1)(x-2)(x-3)}=\dfrac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}$

$\Rightarrow (x)= A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$

Let $x = 1$

$1 = A(1-2)(1-3)+0+0$

$\Rightarrow 1=A\times -1\times -2$

$\Rightarrow A = 1/2$

Again Let $x= 2$

$2 = 0+B(2-1)(2-3)+0$

$\Rightarrow 2=B\times 1\times -1$

$\Rightarrow B = -2$

Again let $x = 3$

$3 =0+ 0+C(3-1)(3-2)$

$\Rightarrow 3=C\times 2\times 1$

$\Rightarrow C = 3/2$

Putting the value of A, B and C in eq (i)

$\dfrac{x}{(x-1)(x-2)(x-3)}=\dfrac{1/2}{(x-1)}+\dfrac{-2}{(x-2)}+\dfrac{3/2}{(x-3)}$

Integrating

$\displaystyle\int\dfrac{3x-1}{(x-1)(x-2)(x-3)}dx=\displaystyle\int\dfrac{1/2}{(x-1)}dx+\displaystyle\int\dfrac{-2}{(x-2)}dx+\displaystyle\int\dfrac{3/2}{(x-3)}dx$

$=\frac{1}{2}\log|x-1|-2\log|x-2|+\frac{3}{2}\log|x-3|+C$

Question 5: $\displaystyle\int \dfrac{2x}{x^2+3x+2}dx$

Solution: $\displaystyle\int\dfrac{2x}{x^2+3x+2}dx=\displaystyle\int\dfrac{2x}{(x+2)(x+1)}dx$

$\dfrac{2x}{(x+2)(x+1)}=\dfrac{A}{(x+2)}+\dfrac{B}{(x+1)}--(i)$

$\Rightarrow \dfrac{2x}{(x+2)(x+1)}=\dfrac{A(x+1)+B(x+2)}{(x+2)(x+1)}$

$\Rightarrow 2x = A(x+1)+B(x+2)$

Let $x=-1$

$2\times -1= A(-1+1)+B(-1+2)$

$\Rightarrow -2 = 0+B$

$\Rightarrow B = -2$

Let $x=-1$

$2\times -2= A(-2+1)+B(-2+2)$

$\Rightarrow -4 = -A+0$

$\Rightarrow A = 4$

Putting in (i) and integrate

$\displaystyle\int \dfrac{2x}{(x+2)(x+1)}dx=\displaystyle\int \dfrac{4}{(x+2)}dx+\displaystyle\int \dfrac{-2}{(x+1)}dx$

$=4\log|x+2|-2\log|x+1|+C$

Question 6: $\displaystyle\int\dfrac{(1-x^2)}{x(1-2x)}dx$

Solution: $\dfrac{(1-x^2)}{x(1-2x)}=\dfrac{1}{2}+\dfrac{A}{x}+\dfrac{B}{1-2x}--(i)$

$\Rightarrow \dfrac{(1-x^2)}{x(1-2x)}=\dfrac{x(1-2x)+A2(1-2x)+B2x}{2x(1-2x)}$

$\Rightarrow 2(1-x^2)=x(1-2x)+A2(1-2x)+B2x$

Let $x=0$

$2(1-0)= 0+2A(1-0)+0$

$\Rightarrow 2 = 2A$

$\Rightarrow A = 1$

Let $x = 1/2$

$2(1-(1/2)^2)=0+2A(1-2\times \frac{1}{2})+2B\times \frac{1}{2}$

$\Rightarrow 2(3/4)=B$

$\Rightarrow B = 3/2$

Putting in eq (i) and integrate

$\displaystyle\int\dfrac{(1-x^2)}{x(1-2x)}dx=\dfrac{1}{2}\displaystyle\int 1dx+\displaystyle\int\dfrac{1}{x}dx+\displaystyle\int\dfrac{3/2}{1-2x}dx$

$=\frac{1}{2}x+\log x+\frac{3}{2}\dfrac{\log|1-2x|}{-2}+C$

$=\frac{1}{2}x+\log x-\frac{3}{4}\log|1-2x|+C$

Question 7: $\displaystyle\int\dfrac{x}{(x^2+1)(x-1)}dx$

Solution: $\dfrac{x}{(x^2+1)(x-1)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}--(i)$

$\Rightarrow \dfrac{x}{(x^2+1)(x-1)}=\dfrac{A(x^2+1)+(Bx+C)(x-1)}{(x^2+1)(x-1)}$

$\Rightarrow x=A(x^2+1)+(Bx+C)(x-1)$

Let $x =1$

$1= A(1+1)+(Bx+C)(1-1)$

$\Rightarrow 1 = A(2)+0$

$\Rightarrow A = 1/2$

Again let $x =0$

$0 = A(0+1)+(B\times 0+C)(0-1)$

$\Rightarrow 0=A-C$

$\Rightarrow A = C=1/2$

Again let $x = -1$

$-1=A((-1)^2+1)+(B\times -1+C)(-1-1)$

$\Rightarrow -1=2A+(-B+C)(-2)$

$\Rightarrow -1=2A+2B-2C$

$\Rightarrow -1=2\times \frac{1}{2}+2B-2\times \frac{1}{2}$

$\Rightarrow -1=1+2B-1$

$\Rightarrow B = -\frac{1}{2}$

Putting in (i) and integrate

$\displaystyle\int\dfrac{x}{(x^2+1)(x-1)}dx=\displaystyle\int\dfrac{1/2}{x-1}dx+\displaystyle\int\dfrac{-1/2x+1/2}{x^2+1}dx$

$=\int\dfrac{1/2}{x-1}dx+\int\dfrac{-1/2x}{x^2+1}dx+\int\dfrac{1/2}{x^2+1}dx$

Let $x^2+1=t$

$\Rightarrow 2x dx=dt$

$\Rightarrow dx=\frac{dt}{2x}$

$=\frac{1}{2}\log|x-1|-\frac{1}{2}\int\frac{1}{t}\frac{dt}{2}+\frac{1}{2}\tan^{-1}x+C$

$=\frac{1}{2}\log|x-1|-\frac{1}{4}\log t+\frac{1}{2}\tan^{-1}x+C$

$=\frac{1}{2}\log|x-1|-\frac{1}{4}\log |x^2+1|+\frac{1}{2}\tan^{-1}x+C$

Question 8: $\displaystyle\int\dfrac{x}{(x-1)^2(x+2)}dx$

Solution: $\dfrac{x}{(x-1)^2(x+2)}dx=\dfrac{A}{(x-1)}+\dfrac{B}{(x-1)^2}+\dfrac{C}{(x+2)}--(i)$

$\Rightarrow \dfrac{x}{(x-1)^2(x+2)}=\dfrac{A(x-1)(x+2)+B(x+2)+C(x-1)^2}{(x-1)^2(x+2)}$

$\Rightarrow x = A(x-1)(x+2)+B(x+2)+C(x-1)^2$

Let $x = 1$

$1 =0+B(1+2)+0$

$\Rightarrow 1 = 3B$

$\Rightarrow B = 1/3$

Let $x = -2$

$-2=0+0+C(-2-1)^2$

$\Rightarrow -2 = 9C$

$\Rightarrow C=-2/9$

Let $x = 0$

$0= A(0-1)(0+2)+B(0+2)+C(0-1)^2$

$\Rightarrow 0 = -2A+2B+C$

$\Rightarrow 0 = -2A+2\times\frac{1}{3}-2/9$

$\Rightarrow \frac{2}{9}-\frac{2}{3}=-2A$

$\Rightarrow -\frac{4}{9}=-2A$

$\Rightarrow A = \frac{2}{9}$

Putting the value of A, B and C in (i) and integrate

$\displaystyle\int\dfrac{x}{(x-1)^2(x+2)}dx=\displaystyle\int\dfrac{2/9}{(x-1)}dx+\displaystyle\int\dfrac{1/3}{(x-1)^2}dx+\displaystyle\int\dfrac{-2/9}{(x+2)}dx$

$=\dfrac{2}{9}\log|x-1|+\dfrac{1}{3}\displaystyle\int(x-1)^{-2}dx-\frac{2}{9}\log|x+2|+C$

$=\dfrac{2}{9}\log\left|\dfrac{x-1}{x+2}\right|+\dfrac{1}{3}\dfrac{(x-1)^{-1}}{(-1)}+C$

$=\dfrac{2}{9}\log\left|\dfrac{x-1}{x+2}\right|-\dfrac{1}{3}\dfrac{1}{(x-1)}+C$

Question 9: $\displaystyle\int\dfrac{(3x+5)}{(x^3-x^2-x+1)}dx$

Solution: Since $x^3-x^2-x+1= x^2(x-1)-1(x-1)$

$=(x-1)(x^2-1)$

$=(x-1)(x-1)(x+1)=(x-1)^2(x+1)$

Thus, $\displaystyle\int\dfrac{(3x+5)}{(x^3-x^2-x+1)}dx=\displaystyle\int\dfrac{(3x+5)}{(x-1)^2(x+1)}dx$

Now, $\dfrac{(3x+5)}{(x-1)^2(x+1)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-1)^2}+\dfrac{C}{(x+1)}--(i)$

$\Rightarrow \dfrac{3x+5}{(x-1)^2(x+1)}=\dfrac{A(x-1)(x+1)+B(x+1)+C(x-1)^2}{(x-1)^2(x+1)}$

$\Rightarrow 3x+5 = A(x-1)(x+1)+B(x+1)+C(x-1)^2$

Let $x = 1$

$3\times 1+5 =0+B(1+1)+0$

$\Rightarrow 8 = 2B$

$\Rightarrow B = 4$

Let $x = -1$

$3\times -1+5=0+0+C(-1-1)^2$

$\Rightarrow 2 = 4C$

$\Rightarrow C=1/2$

Let $x = 0$

$5= A(0-1)(0+1)+B(0+1)+C(0-1)^2$

$\Rightarrow 5 = -1A+B+C$

$\Rightarrow 5 = -A+4+1/2$

$\Rightarrow 5-4-\frac{1}{2}=-A$

$\Rightarrow \frac{1}{2}=-A$

$\Rightarrow A = -\frac{1}{2}$

Putting the value of A, B and C in (i) and integrate

$\displaystyle\int\dfrac{x}{(x-1)^2(x+1)}dx=\displaystyle\int\dfrac{-1/2}{(x-1)}dx+\displaystyle\int\dfrac{4}{(x-1)^2}dx+\displaystyle\int\dfrac{1/2}{(x+1)}dx$

$=-\dfrac{1}{2}\log|x-1|+4\displaystyle\int(x-1)^{-2}dx+\frac{1}{2}\log|x+2|+C$

$=\dfrac{1}{2}\log\left|\dfrac{x+1}{x-1}\right|+4\dfrac{(x-1)^{-1}}{(-1)}+C$

$=\dfrac{1}{2}\log\left|\dfrac{x-1}{x+2}\right|-\dfrac{4}{(x-1)}+C$

Question 10: $\displaystyle\int \dfrac{(2x-3)}{(x^2-1)(2x+3)}dx$

Solution: $\displaystyle\int \dfrac{(2x-3)}{(x^2-1)(2x+3)}dx=\displaystyle\int \dfrac{(2x-3)}{(x-1)(x+1)(2x+3)}dx$

Thus, $\dfrac{(2x-3)}{(x-1)(x+1)(2x+3)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x+1)}+\dfrac{C}{(2x+3)}$

$\Rightarrow \dfrac{(2x-3)}{(x-1)(x+1)(2x+3)}= \dfrac{A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1)}{(x-1)(x+1)(2x+3)}$

$\Rightarrow (2x-3)=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1)$

Let $x=1$

$2\times 1-3 = A(1+1)(2\times 1+3)+0+0$

$\Rightarrow -1=A(2)(5)$

$\Rightarrow -1=10A$

$\Rightarrow A = -1/10$

Let $x=-1$

$2\times -1-3=0+B(-1-1)(2\times -1+3)+0$

$\Rightarrow -5=B(-2)(1)$

$\Rightarrow B=5/2$

Let $x=-3/2$

$2\times -\frac{3}{2}-3=0+0+C(\frac{-3}{2}-1)(\frac{-3}{2}+1)$

$\Rightarrow -6=C(-5/2)(-1/2)$

$\Rightarrow \frac{-24}{5}=C$

Putting the value of A, B and C and integrating

$\displaystyle\int \dfrac{(2x-3)}{(x-1)(x+1)(2x+3)}dx=\displaystyle\int \dfrac{-1/10}{(x-1)}dx+\displaystyle\int \dfrac{5/2}{(x+1)}dx+\displaystyle\int \dfrac{-24/5}{(2x+3)}dx$

$=-\dfrac{1}{10}\log|x-1|+\dfrac{5}{2}\log|x+1|-\dfrac{24}{5}\dfrac{\log|2x+3|}{2}+C$

$=-\dfrac{1}{10}\log|x-1|+\dfrac{5}{2}\log|x+1|-\dfrac{12}{5}\log|2x+3|+C$

Question 11: $\displaystyle\int\dfrac{5x}{(x+1)(x^2-4)}dx$

Solution: $\displaystyle\int\dfrac{5x}{(x+1)(x^2-4)}dx=\int\dfrac{5x}{(x+1)(x-2)(x+2)}$

Now, $\dfrac{5x}{(x+1)(x-2)(x+2)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x-2)}+\dfrac{C}{(x+2)}--(i)$

$\Rightarrow \dfrac{5x}{(x+1)(x-2)(x+2)}=\dfrac{A(x+2)(x-2)+B(x+1)(x+2)+C(x+1)(x-2)}{(x+1)(x-2)(x+2)}$

$\Rightarrow 5x = A(x+2)(x-2)+B(x+1)(x+2)+C(x+1)(x-2)$

Let $x = -1$

$5\times -1=A(-1+2)(-1-2)+0+0$

$\Rightarrow -5 = -3A$

$\Rightarrow A=5/3$

Let $x = 2$

$5\times 2=0+B(2+1)(2+2)+0$

$\Rightarrow 10 = 12B$

$\Rightarrow B=10/12=5/6$

Let $x=-2$

$5\times -2=0+0+C(-2+1)(-2-2)$

$\Rightarrow -10 = 4C$

$\Rightarrow C=-10/4=-5/2$

Putting the value of A,B and C in (i) and integrate

$\displaystyle\int \dfrac{5x}{(x+1)(x-2)(x+2)}dx=\int\dfrac{5/3}{(x+1)}dx+\int\dfrac{5/6}{(x-2)}dx+\int\dfrac{-5/2}{(x+2)}dx$

$=\frac{5}{3}\log |x+1|+\frac{5}{6}\log |x-2|-\frac{5}{2}\log |x+2|+C$

Question 12: $\displaystyle\int\dfrac{x^3+x+1}{x^2-1}dx$

Solution: $\Rightarrow \dfrac{x^3+x+1}{x^2-1}=x+\dfrac{A}{x+1}+\dfrac{B}{x-1}--(i)$

$\Rightarrow \dfrac{x^3+x+1}{x^2-1}=\dfrac{x(x-1)(x+1)+A(x-1)+B(x+1)}{(x+1)(x-1)}$

$\Rightarrow x^3+x+1=x(x-1)(x+1)+A(x-1)+B(x+1)$

Let $x=1$

$1+1+1 = 0+0+B(1+1)$

$\Rightarrow 3=2B$

$\Rightarrow B = 3/2$

Let $x=-1$

$-1-1+1 = 0+A(-1-1)+0$

$\Rightarrow -1=-2A$

$\Rightarrow A = 1/2$

Putting the value of A and B in (i) and integrate

$\displaystyle\int\dfrac{x^3+x+1}{x^2-1}dx=\int xdx+\int \dfrac{1/2}{x+1}dx+\int \dfrac{3/2}{x-1}dx+C$

$= \dfrac{x^2}{2}+\dfrac{1}{2}\log|x+1|+\dfrac{3}{2}\log|x-1|+C$

Question 13: $\displaystyle\int\dfrac{2}{(1-x)(1+x^2)}dx$

Solution: $\dfrac{2}{(1-x)(1+x^2)}=\dfrac{A}{1-x}+\dfrac{Bx+C}{1+x^2}--(i)$

$\Rightarrow \dfrac{2}{(1-x)(1+x^2)}=\dfrac{A(1+x^2)+(Bx+C)(1-x)}{(1-x)(1+x^2)}$

$\Rightarrow 2 = A(1+x^2)+(Bx+C)(1-x)$

Let $x=1$

$2 =A(1+1)+0$

$\Rightarrow 2= 2A$

$\Rightarrow A = 1$

Let $x=0$

$2=A(1+0)+(0+C)(1-0)$

$\Rightarrow 2=A+C$

$\Rightarrow 2 = 1 + C$

$\Rightarrow C = 1$

Let $x = -1$

$2 = A(1+1)+(B\times -1+C)(1+1)$

$\Rightarrow 2 =2A-2B+2C$

$\Rightarrow 2 =2\times 1-2B+2\times 1$

$\Rightarrow 2 = 4-2B$

$\Rightarrow 2B = 4-2$

$\Rightarrow B =1$

Putting the value of A,B and C and integrate

$\displaystyle\int\dfrac{2}{(1-x)(1+x^2)}dx=\int\dfrac{1}{1-x}dx+\int\dfrac{1x+1}{1+x^2}dx$

$=\displaystyle\int\dfrac{1}{1-x}dx+\int \frac{x}{1+x^2}dx+\int \frac{1}{1+x^2}dx$

Let $1+x^2=t$

$\Rightarrow 2x dx=dt$

$\Rightarrow x dx =\frac{dt}{2}$

Now,

$= -\log |1-x|+\displaystyle \int \dfrac{1}{t}\dfrac{dt}{2}+\tan^{-1}x+C$

$= -\log |1-x|+\dfrac{1}{2}\displaystyle \int \dfrac{1}{t}dt+\tan^{-1}x+C$

$= -\log |1-x|+\dfrac{1}{2}\log|t|+\tan^{-1}x+C$

$= -\log |1-x|+\dfrac{1}{2}\log|1+x^2|+\tan^{-1}x+C$

Question 14: $\displaystyle \int \dfrac{3x-1}{(x+2)^2}dx$

Solution: $\dfrac{3x-1}{(x+2)^2}=\dfrac{A}{(x+2)}+\dfrac{B}{(x+2)^2}--(i)$

$\Rightarrow \dfrac{3x-1}{(x+2)^2}=\dfrac{A(x+2)+B}{(x+2)^2}$

$\Rightarrow 3x-1 =A(x+2)+B$

Let $x=-2$

$3\times -2-1=0+B$

$\Rightarrow -7 =B$

Let $x=0$

$3\times 0-1=A(0+2)+B$

$\Rightarrow -1 = 2A+B$

$\Rightarrow -1 = 2A-7$

$\Rightarrow 2A = 6$

$\Rightarrow A = 3$

Putting the value of A and B, and integrate

$\displaystyle\int\dfrac{3x-1}{(x+2)^2}dx=\int\dfrac{3}{(x+2)}dx+\int\dfrac{-7}{(x+2)^2}dx$

$=3\log |x+2|-7\displaystyle\int(x+2)^{-2}dx$

$=3\log |x+2|-7\dfrac{(x+2)^{-2+1}}{-2+1}+C$

$=3\log |x+2|+\dfrac{7}{(x+2)}+C$

Question 15: $\displaystyle\int \dfrac{1}{x^4-1}dx$

Solution:Let $I=\displaystyle\int \dfrac{1}{x^4-1}dx$

$\dfrac{1}{x^4-1}=\dfrac{1}{(x^2-1)(x^2+1)}$

Let $x^2 =y$

$\dfrac{1}{(y-1)(y+1)}=\dfrac{A}{y-1}+\dfrac{B}{y+1}--(i)$

$\Rightarrow \dfrac{1}{(y-1)(y+1)} = \dfrac{A(y+1)+B(y-1)}{(y+1)(y-1)}$

$\Rightarrow 1 = A(y+1)+B(y-1)$

Let $y=1$

$1 = A(1+1)+0$

$\Rightarrow 1 = 2A$

$\Rightarrow A =1/2$

Let $y=-1$

$1 = 0+B(-1-1)$

$\Rightarrow 1 = -2B$

$\Rightarrow B =-1/2$

Puttinig in equation (i)

$\dfrac{1}{(y-1)(y+1)}=\dfrac{1/2}{y-1}+\dfrac{-1/2}{y+1}$

Integrating

$\displaystyle \int \dfrac{1}{(y-1)(y+1)}dx=\int \dfrac{1/2}{y-1}dx+\int \dfrac{-1/2}{y+1}dx$

Replacing $x^2=y$

$\displaystyle \int \dfrac{1}{(x^2-1)(x^2+1)}dx=\int \dfrac{1/2}{x^2-1}dx+\int \dfrac{-1/2}{x^2+1}dx$

$=\dfrac{1}{2}\dfrac{1}{2\times 1}\log\left|\dfrac{x-1}{x+1}\right|-\dfrac{1}{2}\tan^{-1}x+C$

$I=\dfrac{1}{4}\log\left|\dfrac{x-1}{x+1}\right|-\dfrac{1}{2}\tan^{-1}x+C$

Question 16: $\displaystyle\int\dfrac{1}{x(x^n+1)}dx$

Solution: Let $I =\displaystyle\int\dfrac{1}{x(x^n+1)}dx--(i)$

Let $x^n=t$

Differentiate w.r.t. x

$nx^{n-1}dx=dt$

$\Rightarrow dx =\dfrac{dt}{nx^{n-1}}$

Putting in (i)

$I =\displaystyle\int\dfrac{1}{x(t+1)}\dfrac{dt}{nx^{n-1}}$

$=\dfrac{1}{n}\displaystyle\int\dfrac{1}{x^n(t+1)}dt$

$=\dfrac{1}{n}\displaystyle\int\dfrac{1}{t(t+1)}dt$

$=\dfrac{1}{n}\displaystyle\int\dfrac{t+1-t}{t(t+1)}dt$

$=\dfrac{1}{n}\displaystyle\int\dfrac{1}{t}-\dfrac{1}{t+1}dt$

$=\dfrac{1}{n}\left(\log |t|-\log|t+1|\right)+C$

$=\dfrac{1}{n}\log\left|\dfrac{t}{t+1}\right|+C$

$=\dfrac{1}{n}\log\left|\dfrac{x^n}{x^n+1}\right|+C$

Question 17: $\displaystyle \int \dfrac{\cos x}{(1-\sin x)(2-\sin x)}dx$

Solution: Let $I=\displaystyle \int \dfrac{\cos x}{(1-\sin x)(2-\sin x)}dx$

Let $\sin x=t$

Differentiate w.r.t. x

$\Rightarrow \cos xdx =dt$

$I =\displaystyle \int \dfrac{1}{(1-t)(2-t)}dt$

$\dfrac{1}{(1-t)(2-t)}=\dfrac{A}{1-t}+\dfrac{B}{2-t}--(i)$

$\Rightarrow \dfrac{1}{(1-t)(2-t)}= \dfrac{A(2-t)+B(1-t)}{(1-t)(2-t)}$

$\Rightarrow 1 = A(2-t)+B(1-t)$

Let $t=1$

$1 =A(2-1)+0$

$\Rightarrow 1 =A$

Let $t=2$

$2 = 0+B(1-2)$

$\Rightarrow 1= -B$

$\Rightarrow B =-1$

Putting in (i) and integrate

$\displaystyle\int\dfrac{1}{(1-t)(2-t)}dt=\int\dfrac{1}{1-t}dt+\int\dfrac{-1}{2-t}dt$

$=-\log|1-t|+\log |2-t|+C$

$=\log|2-\sin x|-\log|1-\sin x|+C$

$=\log\left|\dfrac{2-\sin x}{1-\sin x}\right|+C$

Question 18: $\displaystyle\int \dfrac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx$

Solution: Let $I =\displaystyle\int \dfrac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx$

Let $x^2=y$

$\dfrac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=\dfrac{(y+1)(y+2)}{(y+3)(y+4)}$

$\dfrac{(y+1)(y+2)}{(y+3)(y+4)}=1+\dfrac{A}{y+3}+\dfrac{B}{y+4}--(i)$

$\Rightarrow \dfrac{(y+1)(y+2)}{(y+3)(y+4)}=\dfrac{(y+3)(y+4)+A(y+4)+B(y+3)}{(y+3)(y+4)}$

$\Rightarrow (y+1)(y+2)=(y+3)(y+4)+A(y+4)+B(y+3)$

Let $y=-3$

$(-3+1)(-3+2)=0+A(-3+4)+0$

$\Rightarrow -2\times -1=A$

$\Rightarrow A =2$

Let $y=-4$

$(-4+1)(-4+2)=0+A(-4+4)+B(-4+3)$

$\Rightarrow -3\times -2=-B$

$\Rightarrow B =-6$

Putting the value of A and B and integrate

$\displaystyle\int\dfrac{(y+1)(y+2)}{(y+3)(y+4)}dx=\int 1dx+\int\dfrac{2}{y+3}dx+\int\dfrac{-6}{y+4}dx$

$\displaystyle\int\dfrac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx=\int 1dx+\int\dfrac{2}{x^2+3}dx+\int\dfrac{-6}{x^2+4}dx$

$=x+2\dfrac{1}{\sqrt{3}}\tan^{-1}\dfrac{x}{\sqrt{3}}-6(\dfrac{1}{2})\tan^{-1}\dfrac{x}{2}+C$

$=x+\dfrac{2}{\sqrt{3}}\tan^{-1}\dfrac{x}{\sqrt{3}}-3\tan^{-1}\dfrac{x}{2}+C$

Question 19: $\displaystyle\int \dfrac{2x}{(x^2+1)(x^2+3)}dx$

Solution: Let $I=\displaystyle\int \dfrac{2x}{(x^2+1)(x^2+3)}dx$

Let $x^2=t$

$\Rightarrow 2x dx =dt$

$I =\displaystyle\int \dfrac{1}{(t+1)(t+3)}dt$

Using partial fraction

$\dfrac{1}{(t+1)(t+3)}=\dfrac{A}{t+1}+\dfrac{B}{t+3}--(i)$

$\Rightarrow \dfrac{1}{(t+1)(t+3)}=\dfrac{A(t+3)+B(t+1)}{(t+1)(t+3)}$

$\Rightarrow 1=A(t+3)+B(t+1)$

Let $t=-1$

$1=A(-1+3)+0$

$\Rightarrow A = 1/2$

Let $t=-3$

$1=A(-3+3)+B(-3+1)$

$\Rightarrow 1 =-2B$

$\Rightarrow B = -1/2$

Putting in eq (i) and integrate

$\displaystyle\int\dfrac{1}{(t+1)(t+3)}dx=\int\dfrac{1/2}{t+1}dx+\int\dfrac{-1/2}{t+3}dx$

$=\dfrac{1}{2}\log|t+1|-\dfrac{1}{2}\log|t+3|+C$

$=\dfrac{1}{2}\log\left|\dfrac{t+1}{t+3}\right|+C$

$=\dfrac{1}{2}\log\left|\dfrac{x^2+1}{x^2+3}\right|+C$

Question 20: $\displaystyle\int \dfrac{1}{x(x^4-1)}dx$

Solution: Let $I=\displaystyle\int \dfrac{1}{x(x^4-1)}dx$

Let $x^4=t$

$\Rightarrow 4x^3 dx=dt$

$\Rightarrow dx =\dfrac{dt}{4x^3}$

$I=\displaystyle\int \dfrac{1}{x(t-1)}\dfrac{dt}{4x^3}$

$\Rightarrow I=\dfrac{1}{4}\displaystyle\int \dfrac{1}{x^4(t-1)}dt$

$\Rightarrow I=\dfrac{1}{4}\displaystyle\int \dfrac{1}{t(t-1)}dt$

Adding and substracting t in numerator and denominator

$\Rightarrow I=\dfrac{1}{4}\displaystyle\int \dfrac{t+1-t}{t(t-1)}dt$

$\Rightarrow I=\dfrac{1}{4}\displaystyle\int \dfrac{1}{(t-1)}-\dfrac{1}{t}dt$

$\Rightarrow I=\dfrac{1}{4}\left[\log|t-1|-\log|t|\right]+C$

$\Rightarrow I=\dfrac{1}{4}\log\left|\dfrac{t-1}{t}\right|+C$

$\Rightarrow I=\dfrac{1}{4}\log\left|\dfrac{x^4-1}{x^4}\right|+C$

Question 21:- $\displaystyle \int\dfrac{1}{e^x-1}dx$

Solution: Let $I= \displaystyle \int\dfrac{1}{e^x-1}dx--(i)$

Let $e^x=t$

$\Rightarrow e^x dx=dt$

$\Rightarrow dx=\dfrac{dt}{e^x}$

Putting in (i)

$I= \displaystyle \int\dfrac{1}{t-1}\dfrac{dt}{e^x}$

$\Rightarrow I= \displaystyle \int\dfrac{1}{t(t-1)}dt$

$\dfrac{1}{t(t-1)}=\dfrac{A}{t}+\dfrac{B}{t-1}$

$\Rightarrow \dfrac{1}{t(t-1)}=\dfrac{A(t-1)+Bt}{t(t-1)}$

$\Rightarrow 1 =A(t-1)+Bt$

Let $t=0$

$1=A(0-1)+0$

$\Rightarrow A = -1$

Let $t=1$

$1=A(1-1)+B$

$\Rightarrow B= 1$

Thus,

$\dfrac{1}{t(t-1)}=\dfrac{-1}{t}+\dfrac{1}{t-1}$

Integrating

$\displaystyle\int\dfrac{1}{t(t-1)}dt=\int\dfrac{-1}{t}dt+\int\dfrac{1}{t-1}dt$

$=-\log|t|+\log|t-1|+C$

$=\log\left|\dfrac{t-1}{t}\right|+C$

$I=\log\left|\dfrac{e^x-1}{e^x}\right|+C$

Choose the correct answer in each of the Exercises 22 and 23.

Question 22: $\int {\dfrac{{xdx}}{{(x - 1)(x - 2)}}}$ equals

A. $A\log \left| {\dfrac{{{{(x - 1)}^2}}}{{x - 2}}} \right| + C$

B. $\log \left| {\dfrac{{{{(x - 2)}^2}}}{{x - 1}}} \right| + C$

C. $\log \left| {{{\left( {\dfrac{{x - 1}}{{x - 2}}} \right)}^2}} \right| + C$

D. $\log |(x - 1)(x - 2)| + C$

Solution: The correct Answer is B.

Let $\dfrac{x}{{(x - 1)(x - 2)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{(x - 2)}}$

$x = A(x - 2) + B(x - 1)$

Let $x=1$

$1=A(1-2)+0$

$\Rightarrow A = -1$

Let $x=2$

$2=A(2-2)+B(2-1)$

$\Rightarrow 2 =B$

$\dfrac{x}{{(x - 1)(x - 2)}} = - \dfrac{1}{{(x - 1)}} + \dfrac{2}{{(x - 2)}}$

$\Rightarrow \int {\dfrac{x}{{(x - 1)(x - 2)}}} dx = \int {\dfrac{{ - 1}}{{(x - 1)}} + \dfrac{2}{{(x - 2)}}} dx$

$= - \log |x - 1| + 2\log |x - 2| + C$

$= \log \left| {\dfrac{{{{(x - 2)}^2}}}{{x - 1}}} \right| + C$

Hence, the correct Answer is B.

Question 23: $\int {\dfrac{{dx}}{{x\left( {{x^2} + 1} \right)}}}$ equals

A. $\log |x| - \dfrac{1}{2}\log \left( {{x^2} + 1} \right) + C$

B. $\log |x| + \dfrac{1}{2}\log \left( {{x^2} + 1} \right) + C$

C. $- \log |x| + \dfrac{1}{2}\log \left( {{x^2} + 1} \right) + C$

D. $\dfrac{1}{2}\log |x| + \log \left( {{x^2} + 1} \right) + C$

Solution : The correct Answer is A.

Let $I =\int {\dfrac{{dx}}{{x\left( {{x^2} + 1} \right)}}}$

Let $x^2=t$

$\Rightarrow 2x dx=dt$

$\Rightarrow dx=\dfrac{dt}{2x}$

$\Rightarrow I=\int {\dfrac{{1}}{{x\left( {{t} + 1} \right)}}}\dfrac{dt}{2x}$

$\Rightarrow I=\dfrac{1}{2}\int {\dfrac{{1}}{{x^2\left( {{t} + 1} \right)}}}dt$

$\Rightarrow I=\dfrac{1}{2}\int {\dfrac{{1}}{{t\left( {{t} + 1} \right)}}}dt$

$\Rightarrow I =\dfrac{1}{2}\int\dfrac{t+1-t}{t(t+1)}dt$

$\Rightarrow I = \dfrac{1}{2}\int\dfrac{1}{t}-\dfrac{1}{t+1}dt$

$\Rightarrow I = \dfrac{1}{2}\left[\log|t|-\log|t+1|\right]+C$

$\Rightarrow I = \dfrac{1}{2}\left[\log|x^2|-\log|x^2+1|\right]+C$

$\Rightarrow I = \dfrac{1}{2}\log|x^2|-\dfrac{1}{2}\log|x^2+1|+C$

$\Rightarrow I = \log|x|-\dfrac{1}{2}\log|x^2+1|+C$

Hence, the correct Answer is A.

https://gmath.in/ex-7-4-integration-ncert-maths-solution-class-12/

EXERCISE 7.5(Integration)

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