The integral dx(1+ x)x-x^2 is equal to: (where C is a constant of integration)

Question

The integral $\displaystyle\int \dfrac{dx}{(1+\sqrt x)\sqrt{x-x^2}}$ is equal to: (where C is a constant of integration)

Accepted Answer

I   =   $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}} $$

=  $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt x \sqrt {1 - x} }}} $$

Let  1 + $$\sqrt x $$ = t

$$ \Rightarrow $$ $\,\,\,$$ ${1 \over {2\sqrt x }}\,dx$$ = dt

I  =  $$\int {{{2dt} \over {t\sqrt {2t - {t^2}} }}} $$

Again let t = $${1 \over z}$$

$$ \Rightarrow $$ $\,\,\,$dt =$-$${1 \over {{z^2}}}dz$$\therefore\,\,\,$I = 2$\int {{{ - {1 \over {{z^2}}}dz} \over {{1 \over z}\sqrt {{2 \over z} - {1 \over {{z^2}}}} }}} $=  2$\int {{{ - dz} \over {\sqrt {2z - 1} }}} $=$ - 2\sqrt {2z - 1}  + C$=$ - 2\sqrt {{2 \over t} - 1}  + C$=$- 2\sqrt {{{2 - t} \over t}}  + C$=$ - 2\sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}}  + C$

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