${\int }_1^2{\mathrm{e}}^{\mathrm{x}}\left(\frac{1}{\mathrm{x}} - \frac{1}{{\mathrm{x}}^2}\right)\mathrm{dx}$ is equal to

• $\mathrm{e} - \frac{{\mathrm{e}}^2}{2}$

• $\frac{{\mathrm{e}}^2}{2} - \mathrm{e}$

• $\frac{{\mathrm{e}}^2}{2} + \mathrm{e}$

• $\frac{{\mathrm{e}}^2}{2} - 2$

The value of ${\int }_{- \mathrm{\pi }}^{\mathrm{\pi }}{\sin }^3\left(\mathrm{x}\right){\cos }^2\left(\mathrm{x}\right)\mathrm{dx}$ is equal to

• 1

• 2

• 3

• 0

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

• $2\sqrt{{\mathrm{x}}^2 + 1} + {\sin }^{-1}\left(\mathrm{x}\right) +\hspace{0.17em}\mathrm{c}$

• $\sqrt{{\mathrm{x}}^2 - 1} - {\sin }^{-1}\left(\mathrm{x}\right) +\hspace{0.17em}\mathrm{c}$

• $2\sqrt{{\mathrm{x}}^2 - 1} + {\sin }^{-1}\left(\mathrm{x}\right) +\hspace{0.17em}\mathrm{c}$

• $\frac{\sqrt{{\mathrm{x}}^2 - 1}}{2} + {\sin }^{-1}\left(\mathrm{x}\right) +\hspace{0.17em}\mathrm{c}$

The value of ${\int }_{- 1}^1\log \left(\frac{\mathrm{x} - 1}{\mathrm{x} +\hspace{0.17em}1}\right)\mathrm{dx}$ is

• 1

• 2

• 0

• 4

345.

Considering four sub-intervals, the value of ${\int }_0^4{2}^{\mathrm{x}}\mathrm{dx}$  by Simpson's rule is

• $\frac{64}{8}$

• $\frac{65}{3}$

• $\frac{62}{12}$

• $\frac{61}{8}$

If I₁ = $\int {\sin }^{-1}\left(\mathrm{x}\right)\mathrm{dx}$ and I₂ = $\int {\sin }^{-1}\left(\sqrt{1 - {\mathrm{x}}^2}\right)\mathrm{dx}$, then

• I₁ = I₂

• ${\mathrm{I}}_2 = \frac{\mathrm{\pi }}{2}{\mathrm{I}}_1$

• ${\mathrm{I}}_1 + {\mathrm{I}}_{2 }= \frac{\mathrm{\pi }}{2}\mathrm{x}$

• ${\mathrm{I}}_1 + {\mathrm{I}}_{2 }= \frac{\mathrm{\pi }}{2}$

$\int \frac{\left(\sin \left(\mathrm{\theta }\right) + \cos \left(\mathrm{\theta }\right)\right)}{\sqrt{\sin \left(2\mathrm{\theta }\right)}}\mathrm{d\theta }$ is equal to

• $\log \left|\cos \left(\mathrm{\theta }\right) - \sin \left(\mathrm{\theta }\right) + \sqrt{\sin \left(2\mathrm{\theta }\right)}\right| + \mathrm{c}$

• $\log \left|\sin \left(\mathrm{\theta }\right) - \cos \left(\mathrm{\theta }\right) + \sqrt{\sin \left(2\mathrm{\theta }\right)}\right| + \mathrm{c}$

• ${\sin }^{-1}\left(\sin \left(\mathrm{\theta }\right) - \cos \left(\mathrm{\theta }\right)\right) + \mathrm{c}$

• ${\sin }^{-1}\left(\sin \left(\mathrm{\theta }\right) + \cos \left(\mathrm{\theta }\right)\right) + \mathrm{c}$

${\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\frac{\mathrm{dx}}{1 + \sqrt{\tan \left(\mathrm{x}\right)}}$ is equal to

• $\frac{\mathrm{\pi }}{12}$

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{6}$

• $\frac{\mathrm{\pi }}{4}$

If f is a continuous function, then

• ${\int }_{- 2}^2\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} = {\int }_0^2\left[\mathrm{f}\left(\mathrm{x}\right) - \mathrm{f}(- \mathrm{x})\right]\mathrm{dx}$

• ${\int }_{- 3}^{5}2\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} = {\int }_{- 6}^{10}\mathrm{f}\left(\mathrm{x} - 1\right)\mathrm{dx}$

• ${\int }_{- 3}^5\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} = {\int }_{- 4}^4\mathrm{f}\left(\mathrm{x} - 1\right)\mathrm{dx}$

• ${\int }_{- 3}^5\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} = {\int }_{- 2}^6\mathrm{f}\left(\mathrm{x} - 1\right)\mathrm{dx}$

$\int \frac{1 + \sin \left(\mathrm{x}\right)}{1 + \cos \left(\mathrm{x}\right)}\mathrm{dx}$ is equal to

• $\mathrm{xtan}\left(\frac{\mathrm{x}}{2}\right) + \mathrm{c}$

• $\log \left(1 + \cos \left(\mathrm{x}\right)\right) +\hspace{0.17em}\mathrm{c}$

• $\cot \left(\frac{\mathrm{x}}{2}\right) +\hspace{0.17em}\mathrm{c}$

• $\log \left(\mathrm{x} + \sin \left(\mathrm{x}\right)\right) +\hspace{0.17em}\mathrm{c}$