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Integrals

Multiple Choice Questions

471.

$\int \frac{dx}{(x + 1) \sqrt{x}}$ is equal to

$\tan^{- 1} (\sqrt{x}) + C$
$2 \tan^{- 1} (x) + C$
$2 \tan^{- 1} (\sqrt{x}) + C$
$\tan^{- 1} (x^{\frac{3}{2}}) + C$

472.

$\int \frac{\log (x)}{x^{2}} dx$ is equal to

$\frac{\log (x)}{x} + \frac{1}{x^{2}} + C$
$- \frac{\log (x)}{x} + \frac{2}{x} + C$
$- \frac{\log (x)}{x} - \frac{1}{2 x} + C$
$- \frac{\log (x)}{x} - \frac{1}{x} + C$

473.

If $\int \frac{f (x)}{\log (\cos (x))} dx = - \log (\log (\cos (x))) + C$ , then f(x) is equal to

$\tan (x)$
- $\sin (x)$
$- \cos (x)$
$- \tan (x)$

474.
$\int \frac{{xsin}^{- 1} (x)}{\sqrt{1 - x^{2}}} dx$ is equal to

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

$x - \sqrt{1 - x^{2}} \sin^{- 1} (x) + C$

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

$x + \sqrt{1 - x^{2}} \sin^{- 1} (x) + C$

$x - \sqrt{1 - x^{2}} \sin^{- 1} (x) + C$

$\int \frac{{xsin}^{- 1} (x)}{\sqrt{1 - x^{2}}} dx Put t = \sin^{- 1} (x) \Rightarrow x = \sin (t) dt = \frac{dx}{\sqrt{1 - x^{2}}} \Rightarrow \int x . t dt = \int \underset{I}{t} . \underset{II}{\sin (t)} = [t (- \cos (t))] - \int 1 . (- \cos (t)) dt = - tcos (t) + \sin (t) + C = - t \sqrt{1 - \sin^{2} (t)} + \sin (t) + C = - \sqrt{1 - x^{2}} \sin^{- 1} (x) + x + C$