Question

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

Answer 1

Answer: 3

$\frac{24}{\pi }\underset{0}{\overset{\sqrt{2}}{\int }}\frac{\left(2-x^2\right)}{\left(x^2+2\right)\sqrt{4+x^4}}dx$
$\frac{24}{\pi }\underset{0}{\overset{\sqrt{2}}{\int }}\frac{x^2\left(\frac{2}{x^2}-1\right)dx}{x\left(x+\frac{2}{x}\right)\times x\sqrt{\frac{4}{x^2}+x^2}}$
$\frac{24}{\pi }\underset{0}{\overset{\sqrt{2}}{\int }}\frac{\left(\frac{2}{x^2}-1\right)dx}{\left(x+\frac{2}{x}\right)\sqrt{{\left(x+\frac{2}{x}\right)}^2-4}}$
$x+\frac{2}{x}=t$
$dt=\left(1-\frac{2}{x^2}\right)dx$
$I=-\frac{24}{\pi }\int \frac{dt}{t\sqrt{t^2-4}}$
$=-\frac{24}{\pi }\times {\frac{1}{2}{\sec }^{-1}\left(\frac{x+\frac{2}{x}}{2}\right)|}_0^{\sqrt{2}}$
$=-\frac{12}{\pi }\left[{\sec }^{-1}\left(\frac{2\sqrt{2}}{2}\right)-{\sec }^{-1}(\infty )\right]$
$=-\frac{12}{\pi }\left[\frac{\pi }{4}-\frac{2\pi }{2\times 2}\right]=-\frac{12}{\pi }\left[-\frac{\pi }{4}\right]$
$=3$