Question

Evaluate: $\int \frac{x^2}{(x^2+2)(2x^2+1)}dx$

Answer 1

$\int \frac{x^2}{(x^2+2)(2x^2+1)}dx$
Let $\frac{x^2}{(x^2+2)(2x^2+1)}=\frac{A}{y+2}+\frac{B}{2y+1}$
$y=A(2y+1)+B(y+2)$
Put $y+2=0\Rightarrow$ $y=-2$
$-2=A(-3)+0$
$\therefore$ $A=\frac{2}{3}$
Put $2y+1=0\Rightarrow$ $y=-\frac{1}{2}$
$-\frac{1}{2}=0+B(-\frac{1}{2}+2)$
$-\frac{1}{2}=B\left(\frac{3}{2}\right)$
$\therefore B=-\frac{1}{3}$
$\therefore \frac{y}{(y+2)(2y+1)}=\frac{2}{3(y+2)}-\frac{1}{3(2y+1)}$
Then $\frac{x^2}{(x^2+2)(2x^2+1)}=\frac{2}{3(x^2+2)}-\frac{1}{3(2x^2+1)}$
$\int \frac{x^2}{(x^2+2)(2x^2+1)}dx=\int \frac{2}{3(x^2+2)}dx-\int \frac{dx}{3(2x^2+1)}$
$=\frac{2}{3}\int \frac{dx}{x^2+{\left(\sqrt{2}\right)}^2}-\frac{1}{3}\int \frac{dx}{2x^2+1}$
$=\frac{2}{3}\times \frac{1}{\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}-\frac{1}{3\times 2}\int \frac{dx}{x^2+{\left(\frac{1}{\sqrt{2}}\right)}^2}$
$=\frac{\sqrt{2}}{3}{\tan }^{-1}\frac{x}{\sqrt{2}}-\frac{1}{6}\times \frac{1}{1/\sqrt{2}}{\tan }^{-1}\frac{x}{1/\sqrt{2}}+c$
$=\frac{\sqrt{2}}{3}{\tan }^{-1}\frac{x}{\sqrt{2}}-\frac{1}{3\sqrt{2}}{\tan }^{-1}\sqrt{2}x+c$