Question

Evaluate $\int \frac{x^2+1}{{(x-1)}^2(x+3)}dx$.

• A. $\frac{5\ln (|x+3|)}{8}-\frac{1}{2(x-1)}+\frac{3\ln (|x-1|)}{8}$
• B. $\frac{5\ln (|x+3|)}{8}+\frac{1}{2(x-1)}+\frac{3\ln (|x-1|)}{8}$
• C. $\frac{5\ln (|x+3|)}{8}-\frac{1}{2(x-1)}-\frac{3\ln (|x-1|)}{8}$
• D. None of these

Answer 1

Answer: (A)

$\frac{5\ln (|x+3|)}{8}-\frac{1}{2(x-1)}+\frac{3\ln (|x-1|)}{8}$

$\int \frac{x^2+1}{{\left(x-1\right)}^2\left(x+3\right)}dx$

$\frac{x^2+1}{{\left(x-1\right)}^2\left(x+3\right)}=\frac{A}{x-1}+\frac{B}{{\left(x-1\right)}^2}+\frac{C}{x+3}$

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

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

$\Rightarrow x^2+1=\left(A+C\right)x^2+\left(2A+b-2C\right)x-3AC+3B$

On Camparing both side we get

$A+C=1$ $2A+B-2C=0$ $-3A+3B+C=1$

$\Rightarrow A=1-C$

from (i) & (ii)

$2A+B-2C=0$

$\Rightarrow 2(1-C)+B-2C=0$

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

$\Rightarrow B-4C+2=0$

$\Rightarrow B=4C-2$ -----------(4)

from (4) & (3) & (1)

$-3A+3B+C=1$

$\Rightarrow -3(1-C)+3\left(4\right(-2\left)\right)+C=1$

$\Rightarrow -3+3C+12C-6+C=1$

$\Rightarrow 16C=10$

$\Rightarrow C=\frac{10}{16}=\frac{5}{8}$

From (1)

$A=1-C=1-\frac{10}{16}=\frac{3}{8}$

From (2)

$B=4C-2=4\times \frac{5}{8}-2=\frac{1}{2}$

Therefore

$\int \frac{x^2+1}{{\left(x-1\right)}^2\left(x+3\right)}dx=\int \frac{dx}{x-1}+\frac{1}{2}\int \frac{dx}{{\left(x-1\right)}^2}+\frac{5}{8}\int \frac{1}{x+3}dx$

$=\frac{3}{8}\log \left|x-1\right|+\frac{1}{2}\times \frac{-1}{\left(x-1\right)}+\frac{5}{8}\log \left|x+3\right|+C$

$=\frac{3}{8}\log \left|x-1\right|-\frac{1}{2\left(x-1\right)}+\frac{5\log \left|x+3\right|}{8}+C$