Who could have done it? 2

We can change the variable to $\displaystyle x^{2}$ , then the integral we want is a half of

$\displaystyle \int_{0}^{1} \dfrac{u}{2(2-u)(1+u)+3\sqrt{(2-u)(1+u)}} du$

Change u to 1-u, we get that the same integral is equal to

$\displaystyle \int_{0}^{1} \dfrac{1-u}{2(2-u)(1+u)+3\sqrt{(2-u)(1+u)}} du$

So, the answer is a fourth of

$\displaystyle \int_{0}^{1} \dfrac{1}{2(2-u)(1+u)+3\sqrt{(2-u)(1+u)}} du$

We let $\displaystyle u=\dfrac{1}{2}+\dfrac{3}{2}\sin(\theta)$ . Note that this substitution comes from completing the square $\displaystyle (2-u)(1+u)$

$\displaystyle \int \dfrac{\sec^{2}(\theta/2)}{6} d\theta = \dfrac{\tan(\theta/2)}{3}$

The limit of integration is $\displaystyle \theta = -\arcsin(1/3)$ to $\displaystyle \arcsin(1/3)$

By solving $\displaystyle \dfrac{1}{2\sqrt{2}}=\dfrac{2x}{1-x^{2}}$ , we get

$\displaystyle \dfrac{1}{4}\cdot \dfrac{2}{3} \cdot (3-2\sqrt{2})=\dfrac{3-2\sqrt{2}}{6}$