x 1-x dx is equal to: - Vidyadip

MCQ

$\int\sqrt{\frac{\text{x}}{1-\text{x}}}\text{ dx}$ is equal to:

A
$\sin^{-1}\sqrt{\text{x}}+\text{C}$
B
$\sin^{-1}\Big\{\sqrt{\text{x}}-\sqrt{\text{x}(1-\text{x})}\Big\}+\text{C}$
C
$\sin^{-1}\Big\{\sqrt{\text{x}(1-\text{x})}\Big\}+\text{C}$
✓
$\sin^{-1}\sqrt{\text{x}}-\sqrt{\text{x}(1-\text{x})}+\text{C}$

✓

Answer

Correct option: D.

$\sin^{-1}\sqrt{\text{x}}-\sqrt{\text{x}(1-\text{x})}+\text{C}$

$\text{I}=\int\sqrt{\frac{\text{x}}{1-\text{x}}}\text{ dx}$
$\text{I}=\int\sqrt{\frac{\text{x}}{1-\text{x}}\cdot\frac{\text{x}}{\text{x}}}\text{ dx}$
$\text{I}=\int\frac{\text{x dx}}{\sqrt{\text{x}-\text{x}^2}}$
Consider,
$\text{x}=\text{A}\frac{\text{d}(\text{x}-\text{x}^2)}{\text{dx}}+\text{B}$
$\text{x}=\text{A}(1-2\text{x})+\text{B}$
$\text{x}=-2\text{Ax}+\text{A}+\text{B}$
$-2\text{A}=1$
$\text{A}=\frac{-1}{2}$
$\text{I}=\int\frac{\frac{-1}{2}(1-2\text{x})+\frac{1}{2}}{\sqrt{\text{x}-\text{x}^2}}\text{ dx}$
$\text{I}=\int\Big(\frac{-1}{2}\frac{1-2\text{x}}{\sqrt{\text{x}-\text{x}^2}}+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}\Big)\text{dx}$
$\text{I}=\frac{-1}{2}\times2\sqrt{\text{x}-\text{x}^2}+\frac{1}{2}\int\frac{1}{\sqrt{\text{x}-\text{x}^2}}\text{ dx}$
Second term after completing square method you will get as
$\text{I}=-\sqrt{\text{x}-\text{x}^2}+\sin^{-1}\sqrt{\text{x}}+\text{C}$

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