$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$ $$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$$
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.

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.
