Question

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

Answer 1

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

Using partial fraction

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

$=\frac{-11}{4}\int \frac{1}{x+3}dx+\frac{11}{4}\int \frac{d}{x+1}dx=\frac{S}{2}\int \frac{1}{(x+1{)}^2}dx$

Sub $x+3=4$ Sub $x+1=5$, ds$=$dx sub $(x+1{)}^2=p^2$

$=-\frac{11}{4}log\left(4\right)+\frac{11}{4}log\left(s\right)-\frac{s}{2}\int \frac{1}{p^2}dp$

$=\frac{5}{2p}+\frac{11log\left(5\right)}{4}-\frac{11log\left(4\right)}{4}+$ constant

$=\frac{1}{4}(-11log(a)+\frac{10}{x+1}+11log(x+1\left)\right)+$ constant

Ans: $\frac{1}{4}(\frac{10}{x+1}+11log(x+1)-11log(x+3\left)\right)+$ constant.