Question

$\int \frac{5x}{\left(x+1\right)\left(x^2-4\right)}dx$

Answer 1

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

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

$\frac{5x}{(x+1)(x^2-4)}=\frac{A}{(x+1)}+\frac{B}{(x-2)}+\frac{C}{(x+2)}$

$\frac{5x}{(x+1)(x^2-4)}=\frac{A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)}{(x+1)(x-2)(x+2)}$

By cancelling denominator

$5x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)$

Putting $x=-1$, in ( 1 )

$5\times (-1)=A(-1-2)(-1+2)+B(-1+1)(-1+2)+C(-1+1)(-1-2)$

$-5=A(-3)\left(1\right)$

$\Rightarrow$ $A=\frac{5}{3}$

Similarly, putting $x=2,$ in ( 1 )

$5\left(2\right)=A(2-2)(2+2)+B(2+1)(2+2)+c(2+1)(2-2)$

$10=A\times 0+B\left(3\right)\left(4\right)+C\times 0$

$10=12B$

$\Rightarrow$ $B=\frac{10}{12}=\frac{5}{6}$

Similarly putting $x=-2,$ in ( 1 )

$5(-2)=A(-2-2)(-2+2)+B(-2+1)(-2+2)+C(-2+1)(-2-2)$

$-10=A\times 0+\times 0+C(-1)(-4)$

$-10=4C$

$C=\frac{-10}{4}=\frac{-5}{2}$

Therefore

$\int \frac{5x}{(x+1)(x^2-4)}=\int (\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+2})dx$

$=\frac{5}{3}\int \frac{dx}{x+1}dx+\frac{5}{6}\int \frac{dx}{x-2}dx-\frac{5}{2}\int \frac{dx}{x+2}$

$=\frac{5}{3}log|x+1|-\frac{5}{2}\log |x+2|+\log |x-2|+c$