Evaluation of $\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$ How does one evaluate the following integral?
$$\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$$
This is a homework problem and I have been evaluating this integral for hours yet no success so far. I have tried to rationalize the integrand by multiplying it with
$$\frac{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}$$
but the integrand is getting worse. I have tried to use trigonometric substitutions like $x=\tan\theta$ and $x=\sqrt{2}\sin\theta$, but I cannot rid off the square root form. I have also tried to use hyperbolic trigonometric substitutions but the thing does not get any easier neither also substitution $y=x^2$ nor $y=\sqrt{(2-x^2)(1+x^2)}$. Using integration by parts is almost impossible for this one. I have also tried to use the tricks from this thread, but still did not get anything. No clue is given. My professor said, we must use clever substitutions but I cannot find them. Any idea or hint? Any help would be appreciated. Thanks in advance.
Edit :
The answer I got from my Prof is $\dfrac{3-2\sqrt{2}}{6}$.

Let us first substitute $y = x^{2} + 1$. Then 
\begin{align*}
I
&:= \int_{0}^{1} \frac{x^{3}}{2(2-x^{2})(1+x^{2}) + 3\sqrt{(2-x^{2})(1+x^{2})}} \, dx \\
&= \frac{1}{2} \int_{1}^{2} \frac{y - 1}{2y(3-y) + 3\sqrt{y(3-y)}} \, dy \tag{1}
\end{align*}
Using the substitution $y \mapsto 3-y$, it follows that 
$$ I = \frac{1}{2} \int_{1}^{2} \frac{2-y}{2y(3-y) + 3\sqrt{y(3-y)}} \, dy. \tag{2} $$
Thus adding (1) and (2) dividing by 2, we obtain
\begin{align*} I
&= \frac{1}{4} \int_{1}^{2} \frac{dy}{2y(3-y) + 3\sqrt{y(3-y)}}
 = \frac{1}{12} \left[ \frac{3 - 2\sqrt{(3-y)y}}{2y-3} \right]_{1}^{2}
 = \frac{3-2\sqrt{2}}{6}.
\end{align*}
Of course this may not be a comprehensive answer. My original approach utilized a chain of substitutions:
\begin{align*} I
&= \frac{1}{4} \int_{1}^{2} \frac{dy}{2y(3-y) + 3\sqrt{y(3-y)}} \\
&= \frac{1}{2} \int_{3/2}^{2} \frac{dy}{2y(3-y) + 3\sqrt{y(3-y)}}, \qquad (\because \text{ by symmetry}) \\
&= \frac{1}{2} \int_{0}^{1} \frac{dy}{(9-s^{2}) + 3\sqrt{9-s^{2}}}, \qquad (s = 2x-3) \\
&= \frac{1}{6} \int_{0}^{\arcsin(1/3)} \frac{d\theta}{1+\cos\theta}, \qquad (s = 3\sin\theta) \\
&= \frac{1}{6} \int_{0}^{3-2\sqrt{2}} dt, \qquad (t = \tan(\theta/2)) \\
&= \frac{3-2\sqrt{2}}{6}.
\end{align*}
