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Integrals

Multiple Choice Questions

191.

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

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

192.

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

$\frac{1}{2} + \frac{1}{2} \log |\cos (x) + \sin (x)| + C$
$\frac{x}{2} + \frac{1}{2} \log |\cos (x) - \sin (x)| + C$
$\frac{1}{2} + \frac{1}{2} \log |\cos (x) - \sin (x)| + C$
$\frac{x}{2} + \frac{1}{2} \log |\cos (x) + \sin (x)| + C$

193.

If $\int_{0}^{a} \frac{x^{2} - 1}{1 - x} dx = - \frac{1}{2}$ , then the value of a is equal to

194.

The value of the integral $\int_{0}^{1} x {(1 - x)}^{5} dx$ is equal to

$\frac{1}{6}$
$\frac{1}{7}$
$\frac{6}{7}$
$\frac{1}{42}$

195.

If [x] denotes the greatest integer less than or equal to x, then the value of $\int_{0}^{2} (|x - 2| + [x]) dx$ is equal to

196.
$\int_{0}^{1} {xe}^{- 5 x} dx$ is equal to

$\frac{1}{25} - \frac{6 e^{- 5}}{25}$

$\frac{1}{25} + \frac{6 e^{- 5}}{25}$

$- \frac{1}{25} - \frac{6 e^{- 5}}{25}$

$\frac{1}{25} + \frac{1}{25} e^{- 5}$

$\frac{1}{25} - \frac{6 e^{- 5}}{25}$

$\int_{0}^{1} \underset{I}{x} \underset{II}{e^{- 5 x}} dx = {[\{x (\frac{e^{- 5 x}}{- 5})\}]}_{0}^{1} - \{\int_{0}^{1} 1 . \frac{e^{- 5 x}}{- 5} dx\} = {[- \frac{{xe}^{- 5 x}}{5} - \frac{e^{- 5 x}}{25}]}_{0}^{1} = - \frac{e^{- 5}}{5} - \frac{e^{- 5}}{25} + \frac{1}{25} = - \frac{6 e^{- 5}}{25} + \frac{1}{25} = \frac{1}{25} - \frac{6 e^{- 5}}{25}$