$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$ $$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$$
My Attempt:
$$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt x\sqrt{1+x(1+x)}}$$
Replacing $x$ by $1-x$,we get
$$\int_{0}^{1}\frac{x}{2-x}.\frac{dx}{\sqrt{1-x}\sqrt{1+(1-x)(2-x)}}$$
Then I got stuck. Please help.

The integral can also be found using a self-similar substitution of $u = \dfrac{1 - x}{1 + x}$.
Here we see that $x = \dfrac{1 - u}{1 + u}$ such that $dx = -\dfrac{2}{(1 + u)^2} \, du$. 
Writing the integral as
$$I = -\int \frac{1 - x}{(1 + x) x \sqrt{x + 1 + \frac{1}{x}}} \, dx,$$
if we observe that
$$x + 1 + \frac{1}{x} = \frac{u^2 + 3}{1 - u^2},$$
then under the self-similar substitution the integral becomes
$$I = \int \frac{2u}{\sqrt{(1 - u^2)(3 + u^2)}} \, du,$$
or
$$I = \int \frac{dt}{\sqrt{(1 - t)(3 + t)}},$$
after setting $t = u^2$. In the denominator, if we complete the square then integrate one has
$$I = \int \frac{dt}{\sqrt{2^2 - (t + 1)^2}} = \sin^{-1} \left (\frac{t + 1}{2} \right ) + C,$$
or
$$\int \frac{x - 1}{(x + 1) \sqrt{x^3 + x^2 + x}} \, dx = \sin^{-1} \left [\frac{1}{2} \left (\frac{1 - x}{1 + x} \right )^2 + \frac{1}{2} \right ] + C.$$
