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Integrals

Multiple Choice Questions

221.

$\int \frac{(1 + x) e^{x}}{\sin^{2} ({xe}^{x})} dx$ is equal to

$- \cot (e^{x}) + C$
$\tan ({xe}^{x}) + C$
$\tan (e^{x}) + C$
$- \cot ({xe}^{x}) + C$

222.

$\int \frac{{xe}^{x}}{(1 + x^{2})} dx$ is equal to

$\frac{- e^{x}}{x + 1} + C$
$\frac{e^{x}}{x + 1} + C$
$\frac{{xe}^{x}}{x + 1} + C$
$\frac{- {xe}^{x}}{x + 1} + C$

223.

$\int e^{x} (\sin (x) + 2 \cos (x)) \sin (x) dx$ is equal to

$e^{x} \cos (x) + C$
$e^{x} \sin (x) + C$
$e^{x} \sin^{2} (x) + C$
$e^{x} \sin (2 x) + C$

224.

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

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

225.

$\int \frac{\sqrt{x^{2} - 1}}{x} dx$ is equal to

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

226.

$\int \frac{\sqrt{5 + x^{2}}}{x^{4}} dx$ is equal to

$\frac{1}{15} {(1 + \frac{5}{x^{2}})}^{\frac{3}{2}}$ + C
$\frac{- 1}{15} {(1 + \frac{1}{x^{2}})}^{\frac{3}{2}} + C$
$\frac{- 1}{15} {(1 + \frac{5}{x^{2}})}^{\frac{3}{2}}$
$\frac{1}{15} {(1 + \frac{1}{x^{2}})}^{\frac{3}{2}}$

227.

The value of $\int_{0}^{1} \frac{dx}{e^{x} + e}$ is equal to

$\frac{1}{e} \log (\frac{1 + e}{2})$
$\log (\frac{1 + e}{2})$
$\frac{1}{e} \log (1 + e)$
$\log (\frac{2}{1 + e})$

228.

The value of the integral $\int_{1}^{e} \frac{1 + \log (x)}{3 x} dx$ is equal to

$\frac{1}{4}$
$\frac{1}{2}$
$\frac{3}{4}$
e

229.
The value of the integtral $\int_{0}^{1} \frac{x^{3}}{1 + x^{8}} dx$ is equal to

$\frac{π}{8}$

$\frac{π}{4}$

$\frac{π}{16}$

$\frac{π}{6}$

$\frac{π}{16}$

$Let I = \int_{0}^{1} \frac{x^{3}}{1 + x^{8}} dx Put x^{4} = t \Rightarrow 4 x^{3} dx = dt ∴ I = \int_{0}^{1} \frac{1}{1 + t^{2}} . \frac{1}{4} dt = \frac{1}{4} {[\tan^{- 1} (t)]}_{0}^{1} = \frac{1}{4} [\tan^{- 1} (1) - \tan^{- 1} (0)] = \frac{1}{4} [\frac{π}{4} - 0] = \frac{π}{16}$