# Evaluate : $\int \large\frac{8\: dx}{(x+2)(x^2+4)}$

Toolbox:
• Form of the partial function$\large\frac{Ax+B}{(x^2+a)}+\frac{Cx+D}{(x^2+b)}$
• $\int \large\frac{dx}{x^2+a^2}=\large\frac{1}{a}$$\tan^{-1}(\large\frac{x}{a})$$+c$
Step 1:
$I=\int\large\frac{8dx}{(x+2)(x^2+4)}$
Consider $\large\frac{8}{(x+2)(x^2+4)}$
This can be resolved into partial fractions as
$\large\frac{A}{x+2}+\frac{Bx+C}{x^2+4}$
$\therefore A(x^2+4)+(x+2)(Bx+C)=8$
Let $x=-2$
$\therefore A(4+4)=8$
$\Rightarrow A=1$
Put $x=0$
$4A+2C=8$
$2C=4$
$C=2$
Put $x=1$
$A(5)+3(B+C)=8$
Substituting for $A$ and $C$ we get,
$5+3(B+2)=8$
$11+3B=8$
$\Rightarrow B=-1$
$A=1,B=-1$ and $C=2$
$\therefore \large\frac{8}{(x+2)(x^2+4)}=\frac{1}{x+2}+\frac{-x+2}{x^2+4}$
Step 2:
$I=\int\large\frac{1}{x+2}-\int\large\frac{x+2}{x^2+4}$
$\;\;=\int\large\frac{dx}{x+2}-\int\large\frac{xdx}{x^2+4}-\int\large\frac{2dx}{x^2+4}$
$\int \large\frac{dx}{x+2}$$=\log\mid x+2\mid------(1) Consider \int\large\frac{x}{x^2+4}$$dx$
Put $x^2+4=t$
$2xdx=dt$
$xdx=\large\frac{dt}{2}$
$\therefore \int\large\frac{x}{x^2+4}$$dx=\large\frac{1}{2}\int \large\frac{dt}{t}=\frac{1}{2}$$\log\mid t\mid$
$\qquad\qquad\;\;\;=\large\frac{1}{2}$$\log(x^2+4)--------(2) Consider \int\large\frac{2}{x^2+4}$$dx=2\int\large\frac{dx}{x^2+4}$
$\int \large\frac{dx}{x^2+a^2}=\large\frac{1}{a}$$\tan^{-1}(\large\frac{x}{a})$$+c$
$\Rightarrow \large\frac{2}{2}$$\tan^{-1}(\large\frac{x}{2})---------(3) Step 3: Combining equ(1),(2) and (3) we get, I=\log\mid x+2\mid-\large\frac{1}{2}$$\log\mid x^2+4\mid +\tan^{-1}(\large\frac{x}{2})+c$