Question

$\int \frac{3x+2}{(x+1)(x+3)}dx=$

• A. $-\frac{7}{2}ln|x+1|+\frac{1}{2}ln|x+3|+c$
• B. $\frac{1}{2}ln|x+1|-\frac{7}{2}ln|x+3|+c$
• C. $\frac{7}{2}ln|x+1|-\frac{1}{2}ln|x+3|+c$
• D. $-\frac{1}{2}ln|x+1|+\frac{7}{2}ln|x+3|+c$

Answer 1

The correct option is D $-\frac{1}{2}ln|x+1|+\frac{7}{2}ln|x+3|+c$

$\int \frac{3x+2}{(x+1)(x+3)}dx$
Resolving $\frac{3x+2}{(x+1)(x+3)}$ into partial fractions
$\frac{3x+2}{(x+1)(x+3)}=\frac{A}{(x+1)}+\frac{B}{(x+3)}$ .....$\left(1\right)$
$3x+2=A(x+3)+B(x+1)$
When $x=-3\Rightarrow B=\frac{7}{2}$
When $x=-1\Rightarrow A=\frac{-1}{2}$
Substituting these values in $\left(1\right)$, we get
$\frac{3x+2}{(x+1)(x+3)}=\frac{-1}{2(x+1)}+\frac{7}{2(x+3)}$
Integrating both sides, we get
$\Rightarrow \int \frac{3x+2}{(x+1)(x+3)}dx=\frac{-1}{2}\int \frac{1}{(x+1)}dx+\frac{7}{2}\int \frac{1}{(x+3)}dx$
$\int \frac{3x+2}{(x+1)(x+3)}dx=\frac{-1}{2}\log |x+1|+\frac{7}{2}\log |x+3|+C$