Question

The value of the integral $\int \frac{5x}{(x+1)(x^2-4)}dx$ is
(where $m$ is an arbitrary constant)

• A. $\frac{5}{3}\ln |x+1|-\frac{5}{2}\ln |x+2|+\frac{5}{6}\ln |x-2|+m$
• B. $\frac{4}{5}\ln |x+1|+\frac{4}{5}\ln |x+2|+\frac{5}{6}\ln |x-2|+m$
• C. $\frac{5}{3}\ln |x+1|-\frac{5}{2}\ln |x+2|+\frac{5}{4}\ln |x-2|+m$
• D. $\frac{4}{5}\ln |x-1|-\frac{5}{2}\ln |x+2|+\frac{5}{6}\ln |x-2|+m$

Answer 1

The correct option is A $\frac{5}{3}\ln |x+1|-\frac{5}{2}\ln |x+2|+\frac{5}{6}\ln |x-2|+m$

$\frac{5x}{(x+1)(x^2-4)}=\frac{5x}{(x+1)(x+2)(x-2)}$
Let
$\frac{5x}{(x+1)(x+2)(x-2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)}\Rightarrow 5x=A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2)\cdots \left(1\right)$
By putting $x=-1,x=-2$ and $x=2$ and on solving, we obtain:
$A=\frac{5}{3},B=-\frac{5}{2},C=\frac{5}{6}\therefore \frac{5x}{(x+1)(x+2)(x-2)}=\frac{5}{3(x+1)}-\frac{5}{2(x+2)}+\frac{5}{6(x-2)}\Rightarrow \int \frac{5x}{(x+1)(x^2-4)}dx=\frac{5}{3}\int \frac{1}{(x+1)}dx-\frac{5}{2}\int \frac{1}{(x+2)}dx+\frac{5}{6}\int \frac{1}{(x-2)}dx=\frac{5}{3}\ln |x+1|-\frac{5}{2}\ln |x+2|+\frac{5}{6}\ln |x-2|+m$
Where $m$ is an arbitrary constant.