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Integrals

Multiple Choice Questions

431.

$\int \frac{4^{x + 1} - 7^{x - 1}}{28^{x}} dx$ is equal to

$\frac{1}{7 \log_{e} (4)} 4^{- x} - \frac{4}{\log_{e} (7)} 7^{- x} + C$
$\frac{1}{7 \log_{e} (4)} 4^{- x} + \frac{4}{\log_{e} (7)} 7^{- x} + C$
$\frac{4^{- x}}{\log_{e} (7)} - \frac{7^{- x}}{\log_{e} (4)} + C$
$\frac{4^{- x}}{\log_{e} (4)} - \frac{7^{- x}}{\log_{e} (7)} + C$

432.

The value of $\int e^{\tan^{- 1} (x)} \frac{(1 + x + x^{2})}{(1 + x^{2})} dx$

$\tan^{- 1} (x) + c$
$e^{\tan^{- 1} (x)} + 2 x + C$
$e^{\tan^{- 1} (x)} + C$
${xe}^{\tan^{- 1} (x)} + C$

433.

The value of $\int \frac{e^{5 \log_{e} (x)} - e^{4 \log_{e} (x)}}{e^{3 \log_{e} (x)} - e^{2 \log_{e} (x)}} dx$ is

x² + C
$\frac{x^{2}}{2} + C$
$\frac{x^{3}}{3} + C$
$\frac{x}{2} + C$

434.

If $f (x) = \frac{\sin^{- 1} (x)}{\sqrt{1 - x^{2}}} and g (x) = e^{\sin^{- 1} (x)}$ , then $\int f (x) g (x) dx$ is equal to

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

435.

The value of $\int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} dx$ is

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

436.
$\int \cos \{2 \tan^{- 1} (\sqrt{\frac{1 - x}{1 + x}})\} dx$ is equal to

$\frac{1}{8} (x^{2} - 1) + C$

$\frac{x^{2}}{4} + C$

$\frac{x}{2} + C$

$\frac{x^{2}}{2} + C$

$\frac{x^{2}}{2} + C$

$Let I = \int \cos \{2 \tan^{- 1} (\sqrt{\frac{1 - x}{1 + x}})\} dx Put x = \cos (θ) ∴ I = \int \cos \{2 \tan^{- 1} (\sqrt{\frac{1 - \cos (θ)}{1 + \cos (θ)}})\} dx = \int \cos \{2 \tan^{- 1} (\tan (\frac{θ}{2}))\} dx = \int \cos (θ) dx = \int xdx = \frac{x^{2}}{2} + C$