Question

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

• A. $\frac{5}{2(x+1)}-\frac{11\left(\ln \right(|x+3|)-\ln (|x+1|\left)\right)}{4}$
• B. $\frac{5}{2(x+1)}+\frac{11\left(\ln \right(|x+3|)-\ln (|x+1|\left)\right)}{4}$
• C. $\frac{5}{2(x-1)}-\frac{11\left(\ln \right(|x+3|)-\ln (|x+1|\left)\right)}{4}$
• D. None of these

Answer 1

Answer: (A)

$\frac{5}{2(x+1)}-\frac{11\left(\ln \right(|x+3|)-\ln (|x+1|\left)\right)}{4}$

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

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

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

$\Rightarrow 3x-2=\left(A+C\right)x^2+\left(4A+B+2C\right)x+3A+3B+C$

On comparing both sides we get

$A+C=0\Rightarrow A=-C$ ---(i)

$4A+B+2C=3$

$\Rightarrow -4C+B+2C=3$

$\Rightarrow B-2C=3$ ---(ii)

and $3A+3B+C=-2$

$\Rightarrow -3C+3B+C=-2$

$\Rightarrow 3B-2C=-2$ ---(iii)

on subtracting (iii) from (ii) we get

$-2B=5$

$\Rightarrow B=\frac{-5}{2}$

from (ii) $B-2C=3$

$\Rightarrow 2C=B-3$

$\Rightarrow 2C=\frac{-5}{2}-3$

$\Rightarrow c=\frac{-11}{4}$

and $A=-C=\frac{11}{4}$

Now, ${\int }^{}\frac{3x-2}{{\left(x+1\right)}^2\left(x+3\right)}dx=\frac{11}{4}{\int }^{}\frac{dx}{x+1}-\frac{5}{2}{\int }^{}\frac{dx}{{\left(x+1\right)}^2}-\frac{11}{4}{\int }^{}\frac{dx}{x+3}$

$=\frac{11}{4}\ln \left|x+1\right|+\frac{5}{2\left(x+1\right)}-\frac{11}{4}\ln \left|x+3\right|+c$

$=\frac{5}{2\left(x+1\right)}-\frac{11\left(\ln \left|x+3\right|-\ln \left|x+1\right|\right)}{4}+c$