Question

The integral
$\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{2 - x^2}{(2 + x^2) \sqrt{4 + x^4}} \, dx$
is equal to _______.

Answer 1

The correct answer is 3
$I = \frac{24}{\pi} \int_{0}^{\sqrt2} \frac{2 - x^2}{(2 + x^2) \sqrt{4 + x^4}} \, dx$
Let
$x = \sqrt2t ⇒ dx = \sqrt2dt$
$I = \frac{24}{\pi} \int_{0}^{1} \frac{(2 - 2t^2) \sqrt{2}}{(2 + 2t^2) \sqrt{4 + 4t^4}} \, dt$
$= \frac{12\sqrt{2}}{\pi} \int_{0}^{1} \frac{\left(\frac{1}{t^2} - 1\right)dt}{\left(t + \frac{1}{t}\right) \sqrt{\left(t + \frac{1}{t}\right)^2 - 2}} \,$
Let
$t + \frac{1}{t} = u$
$⇒ ( 1 - \frac{1}{t²} ) dt = du$
$I = \frac{12\sqrt{2}}{\pi} \int_{2}^{\infty} \frac{-du}{u \sqrt{4^2 - 2}} \, du$
$I = \frac{12\sqrt{2}}{\pi} \int_{2}^{\infty} \frac{du}{u^2 \sqrt{-(\frac{\sqrt{2}}u)^2}}$
$I = \frac{12\sqrt{2}}{\pi} \int_{\frac{1}{\sqrt{2}}}^{0} \frac{-\frac{1}{\sqrt{2}}dp}{\sqrt{1 - p^2}}$
$I = \frac{12}{\pi} \left[ \sin^{-1}(p) \right]_{0}^{\frac{1}{\sqrt{2}}}$
$= \frac{12}{π} . \frac{π}{4}$
= 3