Question

The integral $\frac{24}{\pi} \int\limits_{0}^{\sqrt{2}} \frac{\left(2-x^{2}\right) d x}{\left(2+x^{2}\right) \sqrt{4+x^{4}}}$ is equal to____.

Answer 1

$\frac{24}{\pi} \int\limits_{0}^{\sqrt{2}} \frac{\left(2-x^{2}\right)}{\left(x^{2}+2\right) \sqrt{4+x^{4}}} d x$
$\frac{24}{\pi} \int\limits_{0}^{\sqrt{2}} \frac{x^{2}\left(\frac{2}{x^{2}}-1\right) d x}{x\left(x+\frac{2}{x}\right) \times x \sqrt{\frac{4}{x^{2}}+x^{2}}}$
$\frac{24}{\pi} \int\limits_{0}^{\sqrt{2}} \frac{\left(\frac{2}{x^{2}}-1\right) d x}{\left(x+\frac{2}{x}\right) \sqrt{\left(x+\frac{2}{x}\right)^{2}-4}}$
$x +\frac{2}{ x }= t$
$dt =\left(1-\frac{2}{ x ^{2}}\right) dx$
$I =-\frac{24}{\pi} \int \frac{ dt }{ t \sqrt{ t ^{2}-4}}$
$=-\frac{24}{\pi} \times\left.\frac{1}{2} \sec ^{-1}\left(\frac{x+\frac{2}{x}}{2}\right)\right|_{0} ^{\sqrt{2}}$
$=-\frac{12}{\pi}\left[\sec ^{-1}\left(\frac{2 \sqrt{2}}{2}\right)-\sec ^{-1}(\infty)\right]$
$=-\frac{12}{\pi}\left[\frac{\pi}{4}-\frac{2 \pi}{2 \times 2}\right]=-\frac{12}{\pi}\left[-\frac{\pi}{4}\right]$
$=3$