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

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

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

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

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

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

$Let I = \int \frac{dx}{4 \sin^{2} (x) + 3 \cos^{2} (x)} Dividing numerator : and denominator by \cos^{2} (x), we get = \int \frac{\sec^{2} (x)}{4 \tan^{2} (x) + 3} Put \tan (x) = t \Rightarrow \sec^{2} (x) dx = dt ∴ I = \int \frac{dt}{4 t^{2} + 3} = \frac{1}{4} \int \frac{dt}{t^{2} + \frac{3}{4}} = \frac{1}{4} \times \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} (\frac{1}{\frac{\sqrt{3}}{2}}) + C = \frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 \tan (x)}{\sqrt{3}}) + C$