The integral ∫dx(1+x)x−x2 is equal to (where C is. a constant of integration.)

Let I=∫dx(1+x)x1−xPut 1+x=t⇒12xdx=dt ∴ I=∫2dtt2t−t2Again put t=1z⇒dt=−1z2dz∴ I=2∫−1z2dz1z2z−1z2=2∫−dz2z−1=−22z−1+C=−22t−1+C=−22−tt+C=−21−x1+x+C

The integral ∫dx(1+x)x−x2 is equal to (where C is. a constant of integration.)

Let I=∫dx(1+x)x1−xPut 1+x=t⇒12xdx=dt ∴ I=∫2dtt2t−t2Again put t=1z⇒dt=−1z2dz∴ I=2∫−1z2dz1z2z−1z2=2∫−dz2z−1=−22z−1+C=−22t−1+C=−22−tt+C=−21−x1+x+C

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The integral $\int \frac{d x}{(1 + \sqrt{x}) \sqrt{x - x^{2}}}$ is equal to (where $C$ is. a constant of integration.)

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$- 2 \sqrt{\frac{1 + \sqrt{x}}{1 - \sqrt{x}}} + C$

$- \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} + C$

$- 2 \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} + C$

$2 \sqrt{\frac{1 + \sqrt{x}}{1 - \sqrt{x}}} + C$

answer is C.

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Detailed Solution

Let $I = \int \frac{d x}{(1 + \sqrt{x}) \sqrt{x} \sqrt{1 - x}}$
Put $1 + \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t ∴ I = \int \frac{2 d t}{t \sqrt{2 t - t^{2}}}$
Again put $t = \frac{1}{z} \Rightarrow d t = \frac{- 1}{z^{2}} d z$
$\begin{aligned} ∴ I = 2 \int \frac{- \frac{1}{z^{2}} d z}{\frac{1}{z} \sqrt{\frac{2}{z} - \frac{1}{z^{2}}}} = 2 \int \frac{- d z}{\sqrt{2 z - 1}} = - 2 \sqrt{2 z - 1} + C \\ = - 2 \sqrt{\frac{2}{t} - 1 + C} = - 2 \sqrt{\frac{2 - t}{t}} + C = - 2 \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} + C \end{aligned}$