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Integrals

Multiple Choice Questions

231.

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

$\log |\cos ({xe}^{x})| + C$
$\log |\cot ({xe}^{x})| + C$
$\log |\sec ({xe}^{- x})| + C$
$\log |\sec ({xe}^{x})| + C$

232.

$\int \frac{x^{5}}{\sqrt{1 + x^{3}}} dx$ is equal to

$\frac{2}{9} \sqrt{1 + x^{2}} (x^{3} - 9)$ + C
$\frac{2}{9} \sqrt{x^{3} - 9} (1 + x^{2})$ + C
$\frac{2}{9} \sqrt{1 + x^{3}}$ + C
$\frac{2}{9} \sqrt{1 + x^{2}} (x^{3} - 2)$ + C

233.

$\int \frac{4 e^{x}}{2 e^{x} - 5 e^{- x}} dx$ is equal to

$4 \log |e^{x} - 5| + C$
$\frac{1}{4} \log |e^{2 x} - 5| + C$
$\log |2 e^{2 x} - 5 e^{- x}| + C$
$\log |2 e^{2 x} - 5| + C$

234.

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

$\frac{x^{2}}{2} + 2 x + \log |x| + C$
$\frac{x^{2}}{2} + 2 + \log |x| + C$
$\frac{x^{2}}{2} + x + \log |x| + C$
$\frac{x^{2}}{2} + 2 x + 2 \log |x| + C$

235.

$\int \frac{x^{n - 1}}{x^{2 n} + a^{2}} dx$ is equal to

$\frac{1}{na} \tan^{- 1} (\frac{x^{n}}{a}) + C$
$\frac{n}{a} \sin^{- 1} (\frac{x^{n}}{a}) + C$
$\frac{n}{a} \sin^{- 1} (\frac{x^{n}}{a}) + C$
$\frac{n}{a} \cot^{- 1} (\frac{x^{n}}{a}) + C$

236.

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

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

237.
$\int \frac{1}{xlog (x^{2})} dx$ is equal to

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

$\log |\log (x^{2})| + C$

$2 \log |\log (x^{2})| + C$

$\frac{1}{4} \log |\log (x^{2})| + C$

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

$Let I = \int \frac{1}{xlog (x^{2})} dx Put \log (x^{2}) = t \Rightarrow \frac{1}{x^{2}} . 2 xdx = dt \Rightarrow \frac{2}{x} dx = dt ∴ I = \frac{1}{2} \int \frac{1}{t} dt = \frac{1}{2} \log |t| + C = \frac{1}{2} \log |\log (x^{2})| + C$

238.

$\int_{0}^{1} \frac{1}{(x^{2} + 16) (x^{2} + 25)} dx$ is equal to

$\frac{1}{5} [\frac{1}{4} \tan^{- 1} (\frac{1}{4}) - \frac{1}{5} \tan^{- 1} (\frac{1}{5})]$
$\frac{1}{9} [\frac{1}{4} \tan^{- 1} (\frac{1}{4}) - \frac{1}{5} \tan^{- 1} (\frac{1}{5})]$
$\frac{1}{4} [\frac{1}{4} \tan^{- 1} (\frac{1}{4}) - \frac{1}{5} \tan^{- 1} (\frac{1}{5})]$
$\frac{1}{9} [\frac{1}{5} \tan^{- 1} (\frac{1}{4}) - \frac{1}{4} \tan^{- 1} (\frac{1}{5})]$