Question
Evalute : $\int \frac{2 x+1}{(x+1)(x-2)} d x$

Answer

Let $I =\int \frac{2 x+1}{(x+1)(x-2)} d x$
Let $\frac{2 x+1}{(x+1)(x-2)}=\frac{A}{x+1}+\frac{B}{x-2}$
$
\therefore 2 x +1= A ( x -2)+ B ( x +1)
$
Put $x+1=0$, i.e. $x=-1$, we get
$
\begin{aligned}
& 2(-1)+1= A (-3)+ B (0) \\
& \therefore A =\frac{1}{3}
\end{aligned}
$
Put $x-2=0$, i.e. $x=2$, we get
$
\begin{aligned}
& 2(2)+1= A (0)+ B (3) \\
& \therefore B =\frac{5}{3} \\
& \therefore \frac{2 x+1}{(x+1)(x-2)}=\frac{(1 / 3)}{x+1}+\frac{(5 / 3)}{x-2} \\
& \therefore I=\int\left[\frac{(1 / 3)}{x+1}+\frac{(5 / 3)}{x-2}\right] d x
\end{aligned}
$
$
\begin{aligned}
& =\frac{1}{3} \int \frac{1}{x+1} d x+\frac{5}{3} \int \frac{1}{x-2} d x \\
& =\frac{1}{3} \log |x+1|+\frac{5}{3} \log |x-2|+c .
\end{aligned}
$

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