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Integrals

Multiple Choice Questions

521.

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

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

522.

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

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

523.

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

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

524.

$\int_{- 1}^{1} x (1 - x) (1 + x) dx$ is equal to

$\frac{1}{3}$
$\frac{2}{3}$
1
0

525.

The value of $\int_{- π}^{π} \frac{\sin^{2} (x)}{1 + 7^{x}} dx$ is

7^x
$π$
$\frac{π}{2}$
$2^{π}$

526.
$\int_{0}^{\frac{\sqrt{π}}{2}} 2 x^{3} \sin (x^{2}) dx$ is equal to

$\frac{1}{\sqrt{2}} (1 + \frac{π}{4})$

$\frac{1}{\sqrt{2}} (1 - \frac{π}{4})$

$\frac{1}{\sqrt{2}} (\frac{π}{2} - 1)$

$\frac{1}{\sqrt{2}} (1 - \frac{π}{2})$

$\frac{1}{\sqrt{2}} (1 - \frac{π}{4})$

$I = \int_{0}^{\frac{ππ}{2}} 2 x^{3} \sin (x^{2}) d x = \int_{0}^{\frac{\sqrt{π}}{2}} 2 x^{2} . xsin (x^{2}) dx Put x^{2} = t \Rightarrow 2 xdx = dt Also, when x = 0, then t = 0 and when x = \frac{\sqrt{π}}{2}, then t = \frac{π}{4} \Rightarrow I = \int_{0}^{\frac{π}{4}} \underset{I}{t} \underset{II}{\sin (t)} dt = {[t (- \cos (t))]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} - \cos (t) 1 dt = {[- tcos (t)]}_{0}^{\frac{π}{4}} + \int_{0}^{\frac{π}{4}} \cos (t) dt = {[- tcos (t) + \sin (t)]}_{0}^{\frac{π}{4}} = [- \frac{π}{4} . \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}] = \frac{1}{\sqrt{2}} (1 - \frac{π}{4})$