The integral ∫ dx (1+√x)√x-x^2 is equal to

$\displaystyle I=\int \frac{dx}{(1+\sqrt{x})\sqrt{x}\sqrt{1-x}}$I=∫(1+x​)x​1−x​dx​Put, $1+\sqrt{x}=t \Rightarrow \frac{1}{2\sqrt{x}}dx=dt$1+x​=t⇒2x​1​dx=dt

And, $1-x=1-{(t-1)}^{2}=2t-{t}^{2}$1−x=1−(t−1)2=2t−t2

$$\therefore I= \int \frac{2dt}{t\sqrt{2t-{t}^{2}}}$$∴I=∫t2t−t2​2dt​

Again put, $t=\frac{1}{z}\Rightarrow dt=\frac{-1}{{z}^{2}}dz$t=z1​⇒dt=z2−1​dz

$$\Rightarrow I=2\int \frac{-\frac{1}{{z}^{2}} dz}{\frac{1}{z} \sqrt{\frac{2}{z}-\frac{1}{{z}^{2}}}}=2\int \frac{-dz}{\sqrt{2z-1}}=-2\sqrt{2z-1}+c$$⇒I=2∫z1​z2​−z21​​−z21​dz​=2∫2z−1​−dz​=−22z−1​+c

Where $c$c is arbitrary constant.

$$=-2\sqrt{\frac{2}{t}-1 }+c$$=−2t2​−1​+c

$$=-2\sqrt{\frac{2-t}{t}}+c$$=−2t2−t​​+c

$$=-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+c$$=−21+x​1−x​​​+c