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Integrals

Multiple Choice Questions

431.

If b > a, then $\int_{a}^{b} \frac{dx}{\sqrt{(x - a) (b - x)}}$ is equal to

$\frac{π}{2}$
$π$
$\frac{π}{2} (b - a)$
$\frac{π}{4} (b - a)$

432.

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

$\frac{- e^{x}}{x + 1} + c$
$\frac{e^{x}}{x + 1} + c$
$\frac{e^{x} (x - 1)}{x + 1} + c$
$\frac{{xe}^{x}}{x + 1} + c$

433.

$\int \csc^{4} (x) dx$ is equal to

$\cot (x) + \frac{\cot^{3} (x)}{3} + C$
$\tan (x) + \frac{\tan^{3} (x)}{3} + C$
$- \cot (x) - \frac{\cot^{3} (x)}{3} + C$
$- \tan (x) - \frac{\tan^{3} (x)}{3} + C$

434.

$\int_{0}^{1000} e^{x - [x]} dx$ is

e¹⁰⁰⁰ - 1
$\frac{e^{1000} - 1}{e - 1}$
1000(e - 1)
$\frac{e - 1}{1000}$

435.

$\int \frac{1 + x + \sqrt{x + x^{2}}}{\sqrt{x} + \sqrt{1 + x}} dx$ is equal to

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

436.

$\int \frac{dx}{x^{2} + 4 x + 13}$ is equal to

$\log (x^{2} + 4 x + 13) + c$
$\frac{1}{3} \tan^{- 1} (\frac{x + 2}{3}) + c$
$\log (2 x + 4) + c$
$\frac{2 x + 4}{{(x^{2} + 4 x + 13)}^{2}} + c$

437.

The value of $\int_{2}^{3} \frac{x + 1}{x^{2} (x - 1)} dx$ is

$\log (\frac{16}{9}) + \frac{1}{6}$
$\log (\frac{16}{9}) - \frac{1}{6}$
$2 \log (2) - \frac{1}{6}$
$\log (\frac{4}{3}) - \frac{1}{6}$

438.

$\int_{0}^{\frac{π}{4}} (\cos (x) - \sin (x)) dx + \int_{\frac{π}{4}}^{\frac{5 π}{4}} (\sin (x) - \cos (x)) dx + \int_{2 π}^{\frac{π}{4}} (\cos (x) - \sin (x)) dx$

is equal to

$\sqrt{2} - 2$
$2 \sqrt{2} - 2$
$3 \sqrt{2} - 2$
$4 \sqrt{2} - 2$

439.

$\int \frac{a^{\frac{x}{2}}}{\sqrt{a^{- x} - a^{x}}} dx$ is equal to

$\frac{1}{\log (a)} \sin^{- 1} (a^{x}) + c$
$\frac{1}{\log (a)} \tan^{- 1} (a^{x}) + c$
$2 \sqrt{a^{- x} - a^{x}} + c$
$\log (a^{x} - 1) + c$

440.
The value of $\int_{0}^{1} \frac{x^{4} + 1}{x^{2} + 1} d x$ is

$\frac{1}{6} (3 - 4 π)$

$\frac{1}{6} (3 π + 4)$

$\frac{1}{6} (3 + 4 π)$

$\frac{1}{6} (3 π - 4)$

$\frac{1}{6} (3 π - 4)$

$Let I = \int_{0}^{1} \frac{x^{4} + 1}{x^{2} + 1} d x = \int_{0}^{1} \frac{x^{4} + 1 - 1 + 1}{x^{2} + 1} d x = \int_{0}^{1} [\frac{x^{4} + 1}{x^{2} + 1} + \frac{2}{x^{2} + 1}] dx = \int_{0}^{1} (x^{2} - 1 + \frac{2}{x^{2} + 1}) dx = {[\frac{x^{3}}{3} - x + 2 \tan^{- 1} (x)]}_{0}^{1} = [\frac{1}{3} - 1 + 2 \tan^{- 1} (1) - 0] = - \frac{2}{3} + 2 . \frac{π}{4} = \frac{1}{6} (3 π - 4)$

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

440. The value of ∫01x4 + 1x2 + 1dx is 163 - 4π 163π + 4 163 + 4π 163π - 4

440.
The value of $\int_{0}^{1} \frac{x^{4} + 1}{x^{2} + 1} d x$ is

$\frac{1}{6} (3 - 4 π)$

$\frac{1}{6} (3 π + 4)$

$\frac{1}{6} (3 + 4 π)$

$\frac{1}{6} (3 π - 4)$