Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

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

$\frac{1}{8} (x^{2} - 1) + C$
$\frac{x^{2}}{4} + C$
$\frac{x}{2} + C$
$\frac{x^{2}}{2} + C$

437.

$\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{\tan^{3} (x)} dx$ is

$\frac{π}{12}$
$\frac{π}{4}$
$\frac{π}{3}$
$\frac{π}{6}$

438.

$\int_{- 1}^{1} \frac{17 x^{5} - x^{4} + 29 x^{3} - 31 x + 1}{x^{2} + 1} dx$ is

439.
If I_n = $\int_{0}^{\frac{π}{4}} \tan^{n} (x) dx$ , then $\frac{1}{I_{3} + I_{5}}$ is

1/4

1/2

1/8

4

$∵ I_{n} = \int_{0}^{\frac{π}{4}} \tan^{n} (x) dx ∴ I_{3} + I_{5} = \int_{0}^{\frac{π}{4}} (\tan^{3} (x) + \tan^{5} (x)) dx = \int_{0}^{\frac{π}{4}} \tan^{3} (x) (1 + \tan^{2} (x)) dx = {[\frac{\tan^{4} (x)}{4}]}_{0}^{\frac{π}{4}} = \frac{\tan^{4} (\frac{π}{4})}{4} = \frac{1}{4} Hence, \frac{1}{I_{3} + I_{5}} = 4$