$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$ $$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$$
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.
 A: Hint:
$$I =\int {\frac{{1 - x}}{{1 + x}}\frac{{dx}}{{\sqrt {x + {x^2} + {x^3}} }}}  = \int {\frac{{1 - x}}{{1 + x}}\frac{{dx}}{{\sqrt x \sqrt {1 + x + {x^2}} }}} $$
Making $u=\sqrt{x}$ gives 
$$I=2\int {\frac{{1 - {u^2}}}{{1 + {u^2}}}\frac{{du}}{{\sqrt {1 + {u^2} + {u^4}} }}}  = 2\int {\frac{{\frac{1}{{{u^2}}} - 1}}{{\frac{1}{u} + u}}\frac{{du}}{{\sqrt {\frac{1}{{{u^2}}} + 1 + {u^2}} }}}  =  - 2\int {\frac{1}{{\frac{1}{u} + u}}\frac{{d(\frac{1}{u} + u)}}{{\sqrt {{{(\frac{1}{u} + u)}^2} - 1} }}}$$
I am sure you can do it now.
A: Let $$\displaystyle I = -\int\frac{x-1}{\left(x+1\right)}\cdot \frac{1}{\sqrt{x^3+x^2+x}}dx = -\int\frac{(x^2-1)}{\left(x+1\right)^2\cdot \sqrt{x^3+x^2+x}}dx$$
So $$\displaystyle = -\int\frac{(x^2-1)}{(x^2+2x+1)\sqrt{x^3+x^2+x}}dx  = -\int\frac{\left(1-x^{-2}\right)}{\left(x+\frac{1}{x}+2\right)\sqrt{x+\frac{1}{x}+1}}dx$$
Now Let $$\displaystyle \left(x+\frac{1}{x}+1\right) = t^2\;,$$ Then $$\displaystyle \left(1-\frac{1}{x^2}\right)dx = 2tdt$$
So we get $$\displaystyle I = -2\int\frac{1}{t^2+1}dt = -2\tan^{-1}(t)+\mathcal{C} = -2\left[\frac{\pi}{2}-\cot^{-1}(t)\right]+\mathcal{C}$$
Above we have used the formula $$\bullet \; \displaystyle \tan^{-1}(x)+\cot^{-1}(x) = \frac{\pi}{2}.$$ 
So we get $$\displaystyle I = \int\frac{1-x}{(1+x)\sqrt{x^3+x^2+x}}dx=  2\cot^{-1}\left(\frac{x^2+x+1}{x}\right)+\mathcal{C'}$$
A: 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
$$\int \frac{x - 1}{(x + 1) \sqrt{x^3 + x^2 + x}} \, dx = \sin^{-1} \left [\frac{1}{2} \left (\frac{1 - x}{1 + x} \right )^2 + \frac{1}{2} \right ] + C.$$
