∫01xe- 5xdx is equal to from Mathematics Integrals

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Integrals

Multiple Choice Questions

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481.
$\int_{0}^{1} {xe}^{- 5 x} dx$ is equal to

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

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

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

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

A.

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

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

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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.

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

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

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485.

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

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

486.

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

$\frac{e^{x}}{x} + C$
${xe}^{x} \log |x| + C$
$e^{x} \log |x| + C$
x(e^x + $\log |x|$ ) + C

487.

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

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

488.

$\int (1 - \tan^{2} (x)) dx$ is equal to

$\tan (x) + C$
$\sec (x) + C$
$2 x - \sec (x) + C$
2x - $\tan (x)$ + C

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489.

The value of $\int_{0}^{6} |x - 3| d x$ is equal to

6
0
12
9

490.

If f(x) = $\int_{2 x}^{\sin (x)} \cos (t^{3}) dt$ , then f'(x) is equal to

$\cos (\sin^{3} (x)) \cos (x) - 2 \cos (8 x^{3})$
$\sin (\sin^{3} (x)) \sin (x) - 2 \sin (8 x^{3})$
$\cos (\cos^{3} (x)) \cos (x) - 2 \cos (x^{3})$
$\cos (\sin^{3} (x)) - \cos (8 x^{3})$

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