Ex 7.5 integration ncert maths solution class 12

EXERCISE 7.5(Integration)

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

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/

integration multiple choice question

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