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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

$π$

$\frac{π}{2}$

$2^{π}$

$\frac{π}{2}$

$I = \int_{- π}^{π} \frac{\sin^{2} (x)}{1 + 7^{x}} dx . . . (i) I = \int_{- π}^{π} \frac{\sin^{2} (- x)}{1 + 7^{- x}} I = \int_{- π}^{π} \frac{\sin^{2} (x)}{1 + \frac{1}{7^{x}}} dx = \int_{- π}^{π} \frac{7^{x} \sin^{2} (x)}{1 + 7^{x}} dx . . . (ii) On adding Eqs . (i) and (ii), we get 2 I = \int_{- π}^{π} \frac{1 + 7^{x}}{1 + 7^{x}} \sin^{2} (x) dx = \int_{- π}^{π} \sin^{2} (x) dx = 2 \int_{0}^{π} \sin^{2} (x) dx [∵ \sin^{2} (x) = \sin^{2} (- x)] = 2 \times 2 \int_{0}^{\frac{π}{2}} \sin^{2} (x) dx = 2 \times 2 \int_{0}^{\frac{π}{2}} \frac{1 - \cos (2 x)}{2} dx = 2 {[x - \frac{\sin (2 x)}{2}]}_{0}^{\frac{π}{2}} \Rightarrow 2 I = 2 \times \frac{π}{2} \Rightarrow 2 I = π \Rightarrow I = \frac{π}{2}$