Have you been Looking for **NCERT Class 12 Exercise 7.5 solutions (Integration using Partial Fractions)** for CBSE exam prep?
This post gives step-by-step, BOARD-style answers using partial fractions and related substitutions.
Before you proceed further we recommend you read the following topics for a better understanding
Partial Fractions
Important Integration Formulae
Question 1: Integrate \displaystyle \int \frac{x}{(x+1)(x+2)} \ dx
Solutions:
Let, I = \int\frac{x}{(x+1)(x+2)} \ dx
\frac{x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}
x=A(x+2)+B(x+1)
x=(A+B)x+(2A+B)
Comparing coefficients:
A+B=1,\quad 2A+B=0\Rightarrow A=-1,\ B=2
\frac{x}{(x+1)(x+2)}=-\frac{1}{x+1}+\frac{2}{x+2}
I =-\int\frac{1}{x+1} \ dx + 2\int\frac{1}{x+2} \ dx
I =-\log|x+1|+2\log|x+2|+C
Final Answer: \boxed{\int\frac{x}{(x+1)(x+2)} \ dx = \log\frac{(x+2)^2}{(x+1)}+C}
Question 2: Integrate \displaystyle \int \frac{1}{x^{2}-9} \ dx
Solutions:
Let, I = \int \frac{1}{x^{2}-9} \ dx
x^{2}-9=(x-3)(x+3)
\frac{1}{x^{2}-9}=\frac{A}{x-3}+\frac{B}{x+3}
1=A(x+3)+B(x-3)
1 = (A + B)x + (3A - 3B)
A+B=0,\quad 3A-3B=1\Rightarrow A=\tfrac{1}{6},\ B=-\tfrac{1}{6}
I =\tfrac{1}{6}\int\frac{1}{x-3} \ dx-\tfrac{1}{6}\int\frac{1}{x+3} \ dx
I =\tfrac{1}{6}\log|x-3|-\tfrac{1}{6}\log|x+3|+C
Final Answer: \boxed{\int \frac{1}{x^{2}-9} \ dx = \frac{1}{6}\log\left|\frac{x-3}{x+3}\right|+C}
Question 3: Integrate \displaystyle \int \frac{3x-1}{(x-1)(x-2)(x-3)} \ dx
Solutions:
Let, I = \int \frac{3x-1}{(x-1)(x-2)(x-3)} \ dx
\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}
3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)
\text{Put }x=1:\ 2=2A\Rightarrow A=1
\text{Put }x=2:\ 5=-B\Rightarrow B=-5
\text{Put }x=3:\ 8=2C\Rightarrow C=4
\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{1}{x-1}-\frac{5}{x-2}+\frac{4}{x-3}
I = \int \frac{1}{x-1} \ dx - \int \frac{5}{x-2} \ dx + \int \frac{4}{x-3} \ dx
I =\log|x-1|-5\log|x-2|+4\log|x-3|+C
Final Answer: \boxed{\int \frac{3x-1}{(x-1)(x-2)(x-3)} \ dx = \ln|x-1|-5\ln|x-2|+4\ln|x-3|+C}
Question 4: Integrate \displaystyle \int \frac{x}{(x-1)(x-2)(x-3)} \ dx
Solutions:
Let, I = \int \frac{x}{(x-1)(x-2)(x-3)} \ dx
\frac{x}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}
x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)
\text{Put }x=1:\ 1=2A\Rightarrow A=\tfrac{1}{2}
\text{Put }x=2:\ 2=-B\Rightarrow B=-2
\text{Put }x=3:\ 3=2C\Rightarrow C=\tfrac{3}{2}
I = \frac{1/2}{x-1}-\frac{2}{x-2}+\frac{3/2}{x-3}
I = \int \frac{1/2}{x-1} \ dx - \int \frac{2}{x-2} \ dx + \int \frac{3/2}{x-3} \ dx
I= \frac{1}{2}\log|x-1|-2\log|x-2|+\frac{3}{2}\log|x-3|+C
Final Answer: \boxed{\int \frac{x}{(x-1)(x-2)(x-3)} \ dx = \frac{1}{2}\log|x-1|-2\log|x-2|+\frac{3}{2}\log|x-3|+C:}
Question 5: Integrate \displaystyle \int \frac{2x}{x^{2}+3x+2} \ dx
Solutions:
Let, I = \int \frac{2x}{x^{2}+3x+2} \ dx
D=x^{2}+3x+2, \quad D'=2x+3
2x=(2x+3)-3
\frac{2x}{x^{2}+3x+2} = \frac{2x+3}{x^{2}+3x+2}-\frac{3}{x^{2}+3x+2}
I = \int \frac{2x+3}{x^{2}+3x+2} \ dx - \int \frac{3}{x^{2}+3x+2} \ dx
Let I_1 = \int \frac{2x+3}{x^{2}+3x+2} \ dx
let x^{2}+3x+2 = t
(2x + 3)dx = dt
I_1 = \int \frac{dt}{t}
I_1 = \log{t} + C
I_1 = \log{|x^{2}+3x+2|} + C
I_1 = \log{|(x+1)(x+2)|} + C
I_1 = \log{|x+1|} + \log{|x+2|} + C
Let I_2 = \int \frac{3}{x^{2}+3x+2} \ dx
\frac{3}{x^{2}+3x+2} = \frac{3}{(x+1)(x+2)}
\frac{3}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}
3 = A(x+2) + B(x+1)
\text{Put }x=-1:\ 3=A\Rightarrow A=3
\text{Put }x=-2:\ 3=-B\Rightarrow B=-3
\frac{3}{(x+1)(x+2)} = \frac{3}{x+1} - \frac{3}{x+2}
I_2 = \int \frac{3}{x+1} \ dx - \int \frac{3}{x+2} \ dx
I_2 = {3}\log{|x+1|} - {3}\log{|x+2|} + C
I = I_1 - I_2
I = \log{|x+1|} + \log{|x+2|} - \big( 3\log{|x+1|} - 3\log{|x+2|} \big) + C
I = 4\log|x+2| - 2\log|x+1| + C
Final Answer: \int \frac{2x}{x^{2}+3x+2} \ dx = 4\log|x+2| - 2\log|x+1| + C
Question 6: Integrate \displaystyle \int \frac{1-x^{2}}{x(1-2x)} \ dx
Solutions:
Let, I = \int \frac{1-x^{2}}{x(1-2x)} \ dx
\frac{1-x^{2}}{x(1-2x)}=P+\frac{A}{x}+\frac{B}{1-2x}
1-x^{2}=P x(1-2x)+A(1-2x)+B x
\text{Expand RHS }=(-2P)x^{2}+(P-2A+B)x+A
\text{Compare: }-1=-2P\Rightarrow P=\tfrac{1}{2},\quad 1=A,\quad 0=P-2A+B
0=\tfrac{1}{2}-2\cdot 1+B\Rightarrow B=\tfrac{3}{2}
\frac{1-x^{2}}{x(1-2x)}=\tfrac{1}{2}+\frac{1}{x}+\frac{3/2}{1-2x}
\int\frac{1-x^{2}}{x(1-2x)},dx=\tfrac{1}{2}x+\ln|x|+\tfrac{3}{2}\int\frac{1}{1-2x},dx +C
\int\frac{1}{1-2x} \ dx=-\tfrac{1}{2}\ln|1-2x|
Final Answer: \boxed{\int \frac{1-x^{2}}{x(1-2x)} \ dx ; \tfrac{x}{2}+\ln|x|-\tfrac{3}{4}\ln|1-2x|+C:}
Question 7: Integrate \displaystyle \int \frac{x}{(x^{2}+1)(x-1)} \ dx
Solution:
Let, I = \int \frac{x}{(x^{2}+1)(x-1)} \ dx
\frac{x}{(x^{2}+1)(x-1)}=\frac{A}{x-1}+\frac{Bx+C}{x^{2}+1}
x=A(x^{2}+1)+(Bx+C)(x-1)
x=(A+B)x^{2}+(-B+C)x+(A-C)
Comparing coefficients
0=A+B,\quad 1=-B+C,\quad 0=A-C
A=C,\ B=-A,\ 1=A+C=2A\Rightarrow A=\frac{1}{2},\ B=-\frac{1}{2},\ C=\frac{1}{2}
\frac{x}{(x^{2}+1)(x-1)}=\frac{1/2}{x-1}+\frac{-\frac{1}{2}x+\frac{1}{2}}{x^{2}+1}
I = \int \frac{1/2}{x-1} \ dx + \int \frac{-\frac{1}{2}x \ dx+ \int \frac{1}{2}}{x^{2}+1} \ dx
I =\frac{1}{2}\ln|x-1|-\frac{1}{2}\int\frac{x}{x^{2}+1} \ dx+\frac{1}{2}\int\frac{1}{x^{2}+1} \ dx +C
\int\frac{x}{x^{2}+1} \ dx=\tfrac{1}{2}\ln(x^{2}+1),\quad \int\frac{1}{x^{2}+1} \ dx=\tan^{-1} x
I = \frac{1}{2}\ln|x-1|-\frac{1}{4}\ln(x^{2}+1)+\frac{1}{2}\tan^{-1} x +C
Final Answer: \boxed{\int \frac{x}{(x^{2}+1)(x-1)} \ dx = \frac{1}{2}\ln|x-1|-\frac{1}{4}\ln(x^{2}+1)+\frac{1}{2}\tan^{-1} x +C }
Question 8: Integrate \displaystyle \int \frac{x}{(x-1)^{2}(x+2)} \ dx
Solution:
I = \int \frac{x}{(x-1)^{2}(x+2)} \ dx
\frac{x}{(x-1)^{2}(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+2}
x=A(x-1)(x+2)+B(x+2)+C(x-1)^{2}
x=A(x^2 + x - 2) + B(x+2) + C(x^2 -2x + 1)
x=x^2(A+C) + x(A+B-2C) +(2B - 2A +C)
Comparing Coefficients 0=A+C, \quad 1=A+B-2C, \quad 0=2B-2A+C
0=A+C\Rightarrow C=-A,\quad 1=3A+B,\quad 0=2B-3A \Rightarrow B=\frac{3}{2}A
3A+B=1\Rightarrow 3A+\frac{3}{2}A=1\Rightarrow \frac{9}{2}A=1\Rightarrow A=\frac{2}{9}
B=\frac{1}{3},\ C=-\frac{2}{9}
I =\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
\int\frac{1}{(x-1)^{2}} \ dx=-\frac{1}{x-1}
I =\frac{2}{9}\log|x-1|-\frac{1}{3}\frac{1}{x-1}-\frac{2}{9}\log|x+2| + C
I =\frac{2}{9}\log|\frac{x-1}{x+2}|-\frac{1}{3}\frac{1}{(x-1)} + C
Final Answer: \boxed{ \int \frac{x}{(x-1)^{2}(x+2)} \ dx =\frac{2}{9}\log \big|\frac{x-1}{x+2}\big|-\frac{1}{3}\frac{1}{(x-1)} + C }
Question 9: Integrate \displaystyle \int \frac{3x+5}{x^{3}-x^{2}-x+1} \ dx
Solutions:
Let, I = \int \frac{3x+5}{x^{3}-x^{2}-x+1} \ dx
x^{3}-x^{2}-x+1=(x-1)^{2}(x+1)
\frac{3x+5}{(x-1)^{2}(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+1}
3x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^{2}
\text{Put }x=1:\ 8=2B\Rightarrow B=4
\text{Put }x=-1:\ 2=-2A\Rightarrow A=-\tfrac{1}{2}
3x+5=A(x^2-1) + B(x+1) + C(x^2-2x+1)
3x+5=x^2(A+C) + x(B-2C) +(B+C-A)
Comparing coefficients of x^2
A+C=0\Rightarrow C=\frac{1}{2}
I =-\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
\int\frac{1}{(x-1)^{2}},dx=-\frac{1}{x-1}
I = -\frac{1}{2}\log|x-1|-\frac{4}{x-1}+\frac{1}{2}\log|x+1|+C
I = \frac{1}{2}\log|\frac{x+1}{x-1}|-\frac{4}{x-1}+C
Final Answer: \boxed{\int \frac{3x+5}{x^{3}-x^{2}-x+1} \ dx = \frac{1}{2}\log|\frac{x+1}{x-1}|-\frac{4}{x-1}+C }
Question 10: Integrate \displaystyle \int \frac{2x-3}{(x^{2}-1)(2x+3)} \ dx
Solutions:
Let, I = \int \frac{2x-3}{(x^{2}-1)(2x+3)} \ dx
(x^{2}-1)(2x+3)=(x-1)(x+1)(2x+3)
\frac{2x-3}{(x^{2}-1)(2x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2x+3}
2x-3=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x^{2}-1)
\text{Put }x=1:\ -1=10A\Rightarrow A=-\frac{1}{10}
\text{Put }x=-1:\ -5=-2B\Rightarrow B=\frac{5}{2}
\text{Put }x=-\frac{3}{2}:\ -6=C\left(\frac{9}{4}-1\right)=\frac{5}{4}C\Rightarrow C=-\frac{24}{5}
\frac{2x-3}{(x^{2}-1)(2x+3)}=\frac{-\frac{1}{10}}{x-1}+\frac{\frac{5}{2}}{x+1}+\frac{-\frac{24}{5}}{2x+3}
I = -\frac{1}{10} \int \frac{1}{x-1} \ dx + \frac{5}{2} \int \frac{1}{x+1} \ dx -\frac{24}{5} \int \frac{1}{2x+3} \ dx
I = -\frac{1}{10} \log|x-1| + \frac{5}{2} \log |x+1| -\frac{12}{5} \log |2x + 3|
Final Answer: \boxed{ \int \frac{2x-3}{(x^{2}-1)(2x+3)} \ dx = \frac{5}{2} \log |x+1| -\frac{1}{10} \log|x-1| -\frac{12}{5} \log |2x + 3| }
Question 11: Integrate \displaystyle \int \frac{5x}{(x+1)(x^{2}-4)} \ dx
Solutions:
I = \int \frac{5x}{(x+1)(x^{2}-4)} \ dx
\frac{5x}{(x+1)(x^{2}-4)}=\frac{5x}{(x+1)(x-2)(x+2)}
\frac{5x}{(x+1)(x^{2}-4)} = \frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+2}
5x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)
\text{Put }x=-1: \ -5 =- 3A \implies A = \frac{5}{3}
\text{Put }x=2: \ 10 =12B \implies B = \frac{5}{6}
\text{Put }x=-2: \ -10 = 4C \implies C = \frac{-5}{2}
\frac{5x}{(x+1)(x^{2}-4)} = \frac{\frac{5}{3}}{x+1}+\frac{\frac{5}{6}}{x-2}-\frac{\frac{5}{2}}{x+2}
I = \frac{5}{3} \int \frac{1}{x+1} \ dx + \frac{5}{6} \int \frac{1}{x-2} \ dx - \frac{5}{2} \int \frac{1}{x+2} \ dx
I = \frac{5}{3} \log|x+1| + \frac{5}{6} \log|x-2| - \frac{5}{2} \log|x+2| + C
Final Answer: \boxed{ \int \frac{5x}{(x+1)(x^{2}-4)} \ dx = \frac{5}{3} \log|x+1| + \frac{5}{6} \log|x-2| - \frac{5}{2} \log|x+2| + C }
Question 12: Integrate \int \frac{x^{3}+x+1}{x^{2}-1} \ dx
Solution:
Let, I = \int \frac{x^{3}+x+1}{x^{2}-1} \ dx
\text{Since degree(numerator)}\ge\text{degree(denominator), do division.}
\frac{x^{3}+x+1}{x^{2}-1}=x+\frac{2x+1}{x^{2}-1}
I = \int \big( x+\frac{2x+1}{x^{2}-1} \big) \ dx
I = \int x \ dx + \int \frac{2x+1}{x^{2}-1} \ dx
I = \int x \ dx + \int \frac{2x+1}{x^{2}-1} \ dx
I = \frac{x^2}{2} + I_1
I_1 = \int \frac{2x+1}{x^{2}-1} \ dx
\frac{2x+1}{x^{2}-1} = \frac{2x + 1}{(x+1)(x-1)}
\frac{2x + 1}{(x+1)(x-1)} = \frac{A}{x+1} + \frac{B}{x-1}
2x + 1 = A(x-1) + B(x+1)
for x =1; \quad 3 = 2B \implies B = \frac{3}{2}
for x =-1; \quad -1 = -2A \implies A = \frac{1}{2}
\frac{2x + 1}{(x+1)(x-1)} = \frac{\frac{1}{2}}{x+1} + \frac{\frac{3}{2}}{x-1}
I_1 = \int \frac{1}{2}\frac{1}{x+1} + \frac{3}{2} \frac{1}{x-1} \ dx
I_1 = \frac{1}{2} \int \frac{1}{x+1} \ dx + \frac{3}{2} \int \frac{1}{x-1} \ dx
I_1 = \frac{1}{2} \log{x+1} + \frac{3}{2} \log{x-1}
Substituting back in I
I = \frac{x^2}{2} + -\frac{1}{2} \log{|x+1|} + \frac{3}{2} \log{|x-1|} + C
Final Answer: \boxed { I = \frac{x^2}{2} + -\frac{1}{2} \log{|x+1|} + \frac{3}{2} \log{|x-1|} + C }
Question 13: Integrate \int \frac{2}{(1-x)(1+x^{2})} \ dx
Solutions
Let I = \int \frac{2}{(1-x)(1+x^{2})} \ dx
\frac{2}{(1-x)(1+x^{2})}=\frac{A}{1-x}+\frac{Bx+C}{1+x^{2}}
2=A(1+x^{2})+(Bx+C)(1-x)
2=A+Ax^{2}+Bx(1-x)+C(1-x)
2=A+Ax^{2}+Bx-Bx^{2}+C-Cx.
2 = x^2(A-B) + x(B - C) + (A+C)
Comparing Coeffecients: A-B=0,\ B-C=0,\ A+C=2
A=B,\ B=C\Rightarrow A=B=C,\text{ and }A+C=2\Rightarrow 2A=2\Rightarrow A=1
\therefore A=B=C=1
\frac{2}{(1-x)(1+x^{2})}=\frac{1}{1-x}+\frac{x+1}{1+x^{2}}
I =\int\frac{1}{1-x} \ dx+\int\frac{x}{1+x^{2}} \ dx+\int\frac{1}{1+x^{2}} \ dx
\int\frac{1}{1-x} \ dx=-\ln|1-x|, \quad \int\frac{x}{1+x^{2}} \ dx=\tfrac{1}{2}\ln(1+x^{2}), \quad \int\frac{1}{1+x^{2}} \ dx=\tan^{-1} x.
I = -\ln|1-x|+\tfrac{1}{2}\ln(1+x^{2})+\tan^{-1} x +C
Final Answer: \boxed{\int \frac{2}{(1-x)(1+x^{2})} \ dx = -\ln|1-x|+\tfrac{1}{2}\ln(1+x^{2})+\tan^{-1} x +C}
Question 14: Integrate \displaystyle \int \frac{3x-1}{(x+2)^{2}} \ dx
Solution:
Let, I = \int \frac{3x-1}{(x+2)^{2}} \ dx
\frac{3x-1}{(x+2)^{2}} = \frac{A}{x+2} + \frac{B}{(x+2)^2}
{3x-1} = A(x+2) + B … (1)
For x = -2; \quad -7 = B
Expanding (1)
3x-1 = Ax + 2A + B
Comparing constants: -1 = 2A + B
Substituting B = -7 \implies -1 = 2A - 7 \implies A = 3
\frac{3x-1}{(x+2)^{2}} = \frac{3}{x+2} + \frac{-7}{(x+2)^2}
I = \int \frac{3}{x+2} - \frac{7}{(x+2)^2}
I = 3 \int \frac{1}{x+2} - 7\frac{1}{(x+2)^2}
I = 3 \log|x+2| + \frac{7}{x+2} + C
Final Answer: \boxed{ I = 3 \log|x+2| + \frac{7}{x+2} + C }
Question 15: Integrate \displaystyle \int \frac{1}{x^{4}-1} \ dx
Soution:
Let I = \int \frac{1}{x^{4}-1} \ dx
x^{4}-1=(x^{2}-1)(x^{2}+1)=(x-1)(x+1)(x^{2}+1)
\text{Use partial fractions: }\frac{1}{x^{4}-1}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^{2}+1}.
1=A(x+1)(x^{2}+1)+B(x-1)(x^{2}+1)+(Cx+D)(x^{2}-1)
\text{Put }x=1:\ 1=2A(2)\Rightarrow A=\frac{1}{4}
\text{Put }x=-1:\ 1=-2B(2)\Rightarrow B=-\frac{1}{4}
Expanding the above expression:
1 = A (x^3 + x + x^2 + 1) + B (x^3 + x - x^2 - 1) + C (x^3 - x) + D (x^2 + 1)
1 = x^3 (A + B + C) + x^2 (A - B + D) + x(A + B - C) + (A - B + D)
Comparing coefficients
A + B + C = 0 \implies \frac{1}{4} - \frac{1}{4} + C = 0 \implies C = 0
A - B + D = 0 \implies \frac{1}{4} + \frac{1}{4} + D = 0 \implies D = -\frac{1}{2}
Hence, A=\frac{1}{4}, \quad B=-\frac{1}{4}, \quad C = 0, \quad D = -\frac{1}{2}
\therefore \frac{1}{x^{4}-1}=\frac{1}{4}\cdot\frac{1}{x-1}-\frac{1}{4}\cdot\frac{1}{x+1}+\frac{0\cdot x+ -\tfrac{1}{2}}{x^{2}+1}
I = \frac{1}{4} \int \frac{1}{x-1} \ dx - \frac{1}{4} \int \frac{1}{x+1} \ dx - \frac{1}{2} \int \frac{1}{x^{2}+1} \ dx
I = \frac{1}{4} \log|{x-1}| - \frac{1}{4} \log |{x+1}| - \frac{1}{2} \tan^{-1}x + C
I = \frac{1}{4}\ln\left|\frac{x-1}{x+1}\right| - \frac{1}{2} \tan^{-1}x + C.
Final Answer: \boxed{ \int \frac{1}{x^{4}-1} \ dx = \frac{1}{4}\ln\left|\frac{x-1}{x+1}\right| - \frac{1}{2} \tan^{-1}x+C}
Question 16: Integrate \int \frac{1}{x(x^{n}+1)} \ dx
Solution:
Let, I = \int \frac{1}{x(x^{n}+1)} \ dx
\text{Multiply top and bottom by }x^{n-1}:
I = \int \frac{x^{n-1}}{x^{n}(x^{n}+1)} \ dx
\text{Substitute }t=x^{n} \Rightarrow dt=nx^{n-1}dx \Rightarrow x^{n-1} dx=\tfrac{dt}{n}.
I = \int\frac{1}{n}\cdot\frac{1}{t(t+1)} \ dt
\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}
\therefore I =\frac{1}{n}\int\left(\frac{1}{t}-\frac{1}{t+1}\right) \ dt
I =\frac{1}{n}\left(\log|t|-\log|t+1|\right)+C]
I =\frac{1}{n}\log|\frac{t}{t+1}|+C]
\text{Back-substitute }t=x^{n}
I = \frac{1}{n}\ln\left|\frac{x^{n}}{x^{n}+1}\right|+C
Final Answer: \boxed{ \int \frac{1}{x(x^{n}+1)} \ dx = \frac{1}{n}\ln\left|\frac{x^{n}}{x^{n}+1}\right|+C}
Question 17: Integrate \int \frac{\cos x}{(1-\sin x)(2-\sin x)} \ dx
Solutions:
Let, I = \int \frac{\cos x}{(1-\sin x)(2-\sin x)} \ dx
\text{Put }t=\sin x\Rightarrow dt=\cos x,dx
I = \int\frac{1}{(1-t)(2-t)},dt
\frac{1}{(1-t)(2-t)}=\frac{A}{1-t}+\frac{B}{2-t}
1=A(2-t)+B(1-t)
1=(2A+B)-(A+B)t
Comparing coefficients
- (A+B)=0\Rightarrow B=-A,
2A+B=1\Rightarrow 2A-A=1\Rightarrow A=1.\Rightarrow B=-1
\frac{1}{(1-t)(2-t)}=\frac{1}{1-t}-\frac{1}{2-t}
I =\int\left(\frac{1}{1-t}-\frac{1}{2-t}\right)dt
I=-\log|1-t|+\log|2-t|+C
\text{Back-substitute }t=\sin x
I = -\log|1 - \sin x| + \log|2 - \sin x| + C
I = \log|\frac{2 - \sin x}{1 - \sin x}| + C
Final Answer: \boxed{\int \frac{\cos x}{(1-\sin x)(2-\sin x)} \ dx = \log \left|\frac{2 - \sin x}{1 - \sin x} \right| + C}
Question 18: Integrate \int \frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)} \ dx
Solution:
Let, I = \int \frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)} \ dx
The degree of numerator is equal to the degree of the denominator
Dividing,
\frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)} = 1+\frac{-4x^{2}-10}{(x^{2}+3)(x^{2}+4)}
I = \int 1+\frac{-4x^{2}-10}{(x^{2}+3)(x^{2}+4)} \ dx
I = \int 1 dx - \int\frac{4x^{2}+ 10}{(x^{2}+3)(x^{2}+4)} \ dx
I = x - I_1
I_1 = \int\frac{4x^{2}+ 10}{(x^{2}+3)(x^{2}+4)} \ dx
Using Partial Fractions
\frac{4x^{2}+ 10}{(x^{2}+3)(x^{2}+4)}=\frac{A}{x^{2}+3}+\frac{B}{x^{2}+4}
4x^{2}+ 10 = A(x^{2}+4)+B(x^{2}+3)
4x^{2}+ 10 =(A+B)x^{2}+4A+3B.
Comparing Coefficients
A+B=4 … (1)
4A+3B=10 … (2)
Solving (1) and (2)
A=-2, B=6.
Therefore, \frac{4x^{2}+ 10}{(x^{2}+3)(x^{2}+4)}=\frac{-2}{x^{2}+3}+\frac{6}{x^{2}+4}
I_1 = \int -\frac{2}{x^{2}+3}+\frac{6}{x^{2}+4} \ dx
I_1 = -2 \int \frac{1}{x^{2}+ (\sqrt{3})^2} \ dx+ 6\int \frac{1}{x^{2}+2^2} \ dx
I_1 = \frac{-2}{\sqrt 3} \tan^{-1} \frac{x}{\sqrt 3} + \frac{6}{2} \tan^{-1} \frac{x}{2} + C
Substituting in I
I = x - (\frac{-2}{\sqrt 3} \tan^{-1} \frac{x}{\sqrt 3} + 3 \tan^{-1} \frac{x}{2}) + C
I = x + \frac{2}{\sqrt 3} \tan^{-1} \frac{x}{\sqrt 3} - 3 \tan^{-1} \frac{x}{2} + C
Final Answer: \boxed{ \int \frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)} \ dx = x + \frac{2}{\sqrt 3} \tan^{-1} \frac{x}{\sqrt 3} - 3 \tan^{-1} \frac{x}{2} + C }
Question 19: Integrate \int \frac{2x}{(x^{2}+1)(x^{2}+3)} \ dx
Solution:
Let, I = \int \frac{2x}{(x^{2}+1)(x^{2}+3)} \ dx
\text{Let }t=x^{2}\Rightarrow dt=2xdx
\text{Then} I =\int\frac{1}{(t+1)(t+3)},dt
\frac{1}{(t+1)(t+3)}=\frac{A}{t+1}+\frac{B}{t+3}
1 = A(t+3)+B(t+1)
1 = (A+B)t+(3A+B)
Comparing Coefficients:
A+B=0 \Rightarrow B=-A
3A+B=1 \Rightarrow 3A-A=1
2A=1\Rightarrow A=\frac{1}{2},\ B=-\frac{1}{2}.
I = \frac{1}{2} \int \frac{1}{t+1} \ dx - \frac{1}{2} \int \frac{1}{t+3} \ dx
I = \frac{1}{2}\log|t+1|-\frac{1}{2}\log|t+3|+C
I = \frac{1}{2} \log {| \frac{t+1}{t+3} |} + C
\text{Back-substitute }t=x^{2}
I = \frac{1}{2} \log {| \frac{x^2+1}{x^2+3} |} + C
Final Answer: \boxed{\int \frac{2x}{(x^{2}+1)(x^{2}+3)} \ dx = \frac{1}{2} \log {| \frac{x^2+1}{x^2+3} |} + C}
Question 20: Integrate \displaystyle \int \frac{1}{x(x^{4}-1)} \ dx
Solution:
Let, I = \int \frac{1}{x(x^{4}-1)} \ dx
x^{4}-1=(x^{2}-1)(x^{2}+1)=(x-1)(x+1)(x^{2}+1)
I = \int \frac{1}{(x-1)(x+1)(x^{2}+1)} \ dx
applying partial fractions:
\frac{1}{x(x-1)(x+1)(x^{2}+1)} = \frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}+\frac{Dx+E}{x^{2}+1}
1 = A(x-1)(x+1)(x^2+1) + Bx(x+1)(x^2+1)+Cx(x-1)(x^2+1)+(Dx+E)x(x+1)(x-1) … (1)
For x = 1: \quad 1 = B(2)(2) \implies B = \frac{1}{4}
For x = -1: \quad 1 = -C(-2)(2) \implies C = \frac{1}{4}
For x = 0: \quad 1 = A(-1)(1)(1) \implies A = -1
Expanding (1)
1 = A(x^2-1)(x^2+1) + B(x^2+x)(x^2+1)+C(x^2-x)(x^2+1)+(Dx+E)x(x^2-1)
1 = A(x^4-1) + B(x^4 + x^2+x^3 + x)+C(x^4 + x^2-x^3 - x)+(Dx+E)(x^3-x)
1 = A(x^4-1) + B(x^4 + x^2+x^3 + x)+C(x^4 + x^2-x^3 - x)+(D(x^4-x^2)+E(x^3-x)
Collecting Coefficients
1 = x^4(A+B+C+D) + x^3(B-C+E) + x^2(B+C-D) + x(B - C -E) +(-A)
Comparing coefficients of x^3: \quad 0 = B-C+E \implies 0 = \frac{1}{4} - \frac{1}{4} + E \implies E = 0
Comparing coefficients of x^2: \quad 0 = B+C-D \implies 0 = \frac{1}{4} + \frac{1}{4} - D \implies D = \frac{1}{2}
Therefore
\frac{1}{x(x-1)(x+1)(x^{2}+1)} = \frac{-1}{x}+\frac{\frac{1}{4}}{x-1}+\frac{\frac{1}{4}}{x+1}+\frac{\frac{1}{2}x}{x^{2}+1}
I = \int -\frac{1}{x}+\frac{1}{4}\frac{1}{x-1}+\frac{1}{4}\frac{1}{x+1}+\frac{1}{2}\frac{x}{x^{2}+1} \ dx
I = -\int \frac{1}{x} \ dx + \frac{1}{4} \int \frac{1}{x-1} \ dx + \int \frac{1}{4}\frac{1}{x+1} \ dx + \int \frac{1}{2}\frac{x}{x^{2}+1} \ dx
I = -\log{x} + \frac{1}{4} \log{|x-1|} + \frac{1}{4} \log{|x+1|} + \frac{1}{4} \log{|x^{2}+1|} + C
I = -\log{x} + \frac{1}{4} \log{|(x-1)(x+2)(x^2+1)|} + C
I = -\frac{1}{4}\log{x^4} + \frac{1}{4} \log{|x^4-1)|} + C
I = \frac{1}{4}\log{ \left| \frac{x^4-1)}{x^4} \right|} + C
Final Answer: \boxed {\int \frac{1}{x(x^{4}-1)} \ dx = \frac{1}{4}\log{ \left| \frac{x^4-1)}{x^4} \right|} + C}
Question 21: Integrate \displaystyle \int \frac{1}{e^{x}-1} \ dx
Solution:
Let, I = \int \frac{1}{e^{x}-1} \ dx
\text{Put }t=e^{x} \Rightarrow dt=e^{x} dx = t dx\Rightarrow dx=\frac{dt}{t}
I =\int\frac{1}{t-1}\cdot\frac{dt}{t}
I =\int\frac{1}{t(t-1)} \ dt
\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}
1=A(t-1)+Bt
1=(A+B)t-A
Comparing coefficients
A+B=0, \Rightarrow A=-B
-A = 1 \Rightarrow A=-1,\ B=1
\frac{1}{t(t-1)}=-\frac{1}{t}+\frac{1}{t-1}
I=-\int\frac{1}{t} \ dt+\int\frac{1}{t-1} \ dt
I =-\log|t|+\log|t-1|+C
I = \log |\frac{t-1}{t}| + C
\text{Back-substitute }t=e^{x}
I = \log |\frac{e^{x}-1}{e^{x}}| + C
Final Answer \boxed{\int \frac{1}{e^{x}-1} \ dx = \log \left| \frac{e^{x}-1}{e^{x}} \right| + C}
Question 22: Integrate \displaystyle \int \frac{x}{(x-1)(x-2)} \ dx
(A) \ln\left|\frac{(x-1)^{2}}{x-2}\right|+C
(B) \ln\left|\frac{(x-2)^{2}}{x-1}\right|+C
(C) \ln\left|\left(\frac{x-1}{x-2}\right)^{2}\right|+C
(D) \ln\big| (x-1)(x-2)\big|+C
Solution:
Let, I=\int\frac{x}{(x-1)(x-2)},dx
\frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}
x=A(x-2)+B(x-1)
x=(A+B)x+(-2A-B)
Comparing Coefficients
-2A-B=0\Rightarrow B=-2A
A+B=1, \Rightarrow A-2A=1\Rightarrow -A=1\Rightarrow A=-1,\quad B=2
\frac{x}{(x-1)(x-2)}=-\frac{1}{x-1}+\frac{2}{x-2}
I=-\int\frac{1}{x-1},dx+2\int\frac{1}{x-2} \ dx
I=-\ln|x-1|+2\ln|x-2|+C
\boxed{ \int\frac{x}{(x-1)(x-2)} \ dx=-\ln|x-1|+2\ln|x-2|+C \ }
Correct choice: (B).
Question 23: Integrate \displaystyle \int \frac{dx}{x(x^{2}+1)}
(A) \ln|x|-\tfrac{1}{2}\ln(x^{2}+1)+C
(B) \ln|x|+\tfrac{1}{2}\ln(x^{2}+1)+C
(C) -\ln|x|+\tfrac{1}{2}\ln(x^{2}+1)+C
(D) \tfrac{1}{2}\ln|x|+\ln(x^{2}+1)+C
Solution:
Let, I=\int\frac{1}{x(x^{2}+1)} \ dx
\frac{1}{x(x^{2}+1)}=\frac{A}{x}+\frac{Bx+C}{x^{2}+1}
1=A(x^{2}+1)+(Bx+C)x=A x^{2}+A+Bx^{2}+Cx
1=(A+B)x^{2}+Cx+A
Comparing coefficients:
A+B=0,\ C=0,\ A=1
\Rightarrow A=1,\ B=-1,\ C=0
\frac{1}{x(x^{2}+1)}=\frac{1}{x}-\frac{x}{x^{2}+1}
I=\int\frac{1}{x},dx-\int\frac{x}{x^{2}+1},dx
I=\log|x|-\frac{1}{2}\ln(x^{2}+1)+C
\boxed{\int\frac{dx}{x(x^{2}+1)}=\log|x|-\frac{1}{2}\log(x^{2}+1)+C}
Correct choice: (A)