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Integrals

Multiple Choice Questions

471.

$\int \frac{dx}{(x + 1) \sqrt{x}}$ is equal to

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

472.

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

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

473.

If $\int \frac{f (x)}{\log (\cos (x))} dx = - \log (\log (\cos (x))) + C$ , then f(x) is equal to

$\tan (x)$
- $\sin (x)$
$- \cos (x)$
$- \tan (x)$

474.

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

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

475.

$\int \frac{4 e^{x} - 6 e^{- x}}{9 e^{x} - 4 e^{- x}} dx$ is equal to

$\frac{3}{2} x + \frac{35}{36} \log |9 e^{2 x} - 4| + C$
$\frac{3}{2} x - \frac{35}{36} \log |9 e^{2 x} - 4| + C$
$- \frac{3}{2} x + \frac{35}{36} \log |9 e^{2 x} - 4| + C$
$- \frac{5}{2} x + \frac{35}{36} \log |9 e^{2 x} - 4| + C$

476.

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

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

477.
$\int \frac{dx}{1 + \tan (x)}$ is equal to

$\frac{1}{2} + \frac{1}{2} \log |\cos (x) + \sin (x)| + C$

$\frac{x}{2} + \frac{1}{2} \log |\cos (x) - \sin (x)| + C$

$\frac{1}{2} + \frac{1}{2} \log |\cos (x) - \sin (x)| + C$

$\frac{x}{2} + \frac{1}{2} \log |\cos (x) + \sin (x)| + C$

$\frac{x}{2} + \frac{1}{2} \log |\cos (x) + \sin (x)| + C$

$\int \frac{dx}{1 + \tan (x)} = \int \frac{\cos (x)}{\sin (x) + \cos (x)} \times \frac{\cos (x) - \sin (x)}{\cos (x) - \sin (x)} dx = \int \frac{(\cos^{2} (x) - \sin (x) \cos (x))}{(\cos^{2} (x) - \sin^{2} (x))} dx = \frac{1}{2} \int \frac{2 \cos^{2} (x)}{\cos (2 x)} dx - \frac{1}{2} \int \frac{2 \sin (x) . \cos (x)}{\cos (2 x)} dx = \frac{1}{2} \int \frac{(1 + \cos (2 x))}{\cos (2 x)} dx - \frac{1}{2} \int \frac{\sin (2 x)}{\cos (2 x)} dx = \frac{1}{2} \int \sec (2 x) dx + \frac{1}{2} \int dx - \frac{1}{2} \int \tan (2 x) dx = \frac{1}{2} . \frac{1}{2} \log |\sec (2 x) + \tan (2 x)| + \frac{1}{2} . x - \frac{1}{2} . \frac{1}{2} \log |\sec (2 x)| + C = \frac{1}{4} \log |\sec (2 x) + \tan (2 x)| - \frac{1}{4} \log |\sec (2 x)| + \frac{x}{2} + C = \frac{x}{2} + \frac{1}{4} \log |\frac{\sec (2 x) + \tan (2 x)}{\sec (2 x)}| + C = \frac{x}{2} + \frac{1}{4} \log |1 + \sin (2 x)| C = \frac{x}{2} + \frac{1}{4} \log |(\sin^{2} (x) + \cos^{2} (x)) + 2 \sin (x) \cos (x)| + C = \frac{x}{2} + \frac{1}{4} \log |{(\sin (x) + \cos (x))}^{2}| + C = \frac{x}{2} + \frac{1}{2} \log |\cos (x) + \sin (x)| + C$

478.

If $\int_{0}^{a} \frac{x^{2} - 1}{1 - x} dx = - \frac{1}{2}$ , then the value of a is equal to

479.

The value of the integral $\int_{0}^{1} x {(1 - x)}^{5} dx$ is equal to

$\frac{1}{6}$
$\frac{1}{7}$
$\frac{6}{7}$
$\frac{1}{42}$

480.

If [x] denotes the greatest integer less than or equal to x, then the value of $\int_{0}^{2} (|x - 2| + [x]) dx$ is equal to

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

477. ∫dx1 + tanx is equal to 12 + 12logcosx + sinx + C x2 + 12logcosx - sinx + C 12 + 12logcosx - sinx + C x2 + 12logcosx + sinx + C

477.
$\int \frac{dx}{1 + \tan (x)}$ is equal to

$\frac{1}{2} + \frac{1}{2} \log |\cos (x) + \sin (x)| + C$

$\frac{x}{2} + \frac{1}{2} \log |\cos (x) - \sin (x)| + C$

$\frac{1}{2} + \frac{1}{2} \log |\cos (x) - \sin (x)| + C$

$\frac{x}{2} + \frac{1}{2} \log |\cos (x) + \sin (x)| + C$