STD 12 - 7.1 inderfinite integral — JEE कक्षा 11 साइन्स — Question

${\rm{ Put }}1 + \sqrt x  = t$

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

$\Rightarrow \quad I=\int \frac{2 d t}{t \sqrt{2 t-t^{2}}}$

Again put $t=\frac{1}{z}$

$ \Rightarrow d t=\frac{-1}{2^{2}} d z$

$ = I = 2\int {\frac{{ - \frac{1}{{{z^2}}}dz}}{{\frac{1}{z}\sqrt {\frac{2}{z} - \frac{1}{{{z^2}}}} }}}  $

$= 2\int {\frac{{ - dz}}{{\sqrt {2z - 1} }}} $

$=-2 \sqrt{2 z-1}+c$

$=-2 \sqrt{\frac{2}{t}-1+c}$

$=-2 \sqrt{\frac{2-t}{t}}+c$

$=-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+c$