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

475.
$\int \frac{4 e^{x} - 6 e^{- x}}{9 e^{x} - 4 e^{- x}} dx$ is equal to

$\frac{3}{2} x + \frac{35}{36} \log |9 e^{2 x} - 4| + C$

$\frac{3}{2} x - \frac{35}{36} \log |9 e^{2 x} - 4| + C$

$- \frac{3}{2} x + \frac{35}{36} \log |9 e^{2 x} - 4| + C$

$- \frac{5}{2} x + \frac{35}{36} \log |9 e^{2 x} - 4| + C$

$- \frac{3}{2} x + \frac{35}{36} \log |9 e^{2 x} - 4| + C$

$\int \frac{4 e^{x} + 6 e^{- x}}{9 e^{2 x} - 4} dx = \int \frac{4 e^{2 x} + 6}{9 e^{2 x} - 4} dx = 4 \int \frac{e^{2 x}}{9 e^{2 x} - 4} + 6 \int \frac{e^{- 2 x}}{9 - 4 e^{- 2 x}} Put t_{1} = 9 e^{2 x} - 4 and t_{2} = 9 - 4 e^{- 2 x} {dt}_{1} = 18 e^{2 x} dx and {dt}_{2} = 8 e^{- 2 x} dx = 4 \int \frac{1}{t_{1}} . \frac{{dt}_{1}}{18} + 6 \int \frac{1}{t_{2}} . \frac{{dt}_{2}}{8} = \frac{2}{9} \int \frac{{dt}_{1}}{t_{1}} + \frac{3}{4} \int \frac{{dt}_{2}}{t_{2}} = \frac{2}{9} \log (t_{1}) + \frac{3}{4} \log (t_{2}) + C = \frac{2}{9} \log |9 e^{2 x} - 4| + \frac{3}{4} \log |9 - 4 e^{- 2 x}| + C = \frac{2}{9} \log |9 e^{2 x} - 4| + \frac{3}{4} \log |9 e^{2 x} - 4| - \frac{3}{4} \log |e^{2 x}| + C = (\frac{2}{9} + \frac{3}{4}) og |9 e^{2 x} - 4| - \frac{3}{4} (2 x) + C [∵ \log (e) = 1] = - \frac{3}{2} x + \frac{35}{36} \log |9 e^{2 x} - 4| + C$