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Integrals

Multiple Choice Questions

481.

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

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

482.

$\int \frac{5 x dx}{{(1 - x)}^{3}}$ is equal to

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

483.

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

$2 \log |\sqrt{x} - 1| + C$
$2 \log |\sqrt{x} + 1| + C$
$\log |\sqrt{x} - 1| + C$
$\frac{1}{2} \log |\sqrt{x} + 1| + C$

484.

$\int \frac{dx}{4 \sin^{2} (x) + 3 \cos^{2} (x)}$

$\frac{\sqrt{3}}{4} \tan^{- 1} (\frac{2 \tan (x)}{\sqrt{3}}) + C$
$\frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{\tan (x)}{\sqrt{3}}) + C$
$\frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 \tan (x)}{\sqrt{3}}) + C$
$\frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 \tan (x)}{\sqrt{3}}) + C$

485.
$\int \frac{\sec (x) dx}{\sqrt{\cos (2 x)}}$ is equal to

2 $\sin^{- 1} (\tan (x))$

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

$\sin^{- 1} (\tan (x))$

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

$\sin^{- 1} (\tan (x))$

$Let I = \int \frac{\sec (x)}{\sqrt{\cos (2 x)}} dx = \int \frac{\sec (x)}{\sqrt{\frac{1 - \tan^{2} (x)}{1 + \tan^{2} (x)}}} dx = \int \frac{\sec^{2} (x)}{\sqrt{1 - \tan^{2} (x)}} Put \tan (x) = t \Rightarrow \sec^{2} (x) dx = dt ∴ I = \int \frac{dt}{\sqrt{1 - t^{2}}} = \sin^{- 1} (t) + C = \sin^{- 1} (\tan (x)) + C$