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

Correct Answer: 3

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