271.

$\int {\mathrm{e}}^{- \log \left(\mathrm{x}\right)}\mathrm{dx}$ is equal to :

• e^{- log(x)} + C

• - xe^{- log(x)} + C

• e^log(x) + C

• $\log \left|\mathrm{x}\right| + \mathrm{C}$

$\int \frac{{\mathrm{a}}^{{\displaystyle \raisebox{1ex}{$\mathrm{x}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\sqrt{{\mathrm{a}}^{- \mathrm{x}} - {\mathrm{a}}^{\mathrm{x}}}}\mathrm{dx}$ is equal to :

• $\frac{1}{\log \left(\mathrm{a}\right)}{\sin }^{-1}\left({\mathrm{a}}^{\mathrm{x}}\right) + \mathrm{c}$

• $\frac{1}{\log \left(\mathrm{a}\right)}{\tan }^{-1}\left({\mathrm{a}}^{\mathrm{x}}\right) + \mathrm{c}$

• $2\sqrt{{\mathrm{a}}^{- \mathrm{x}} - {\mathrm{a}}^{\mathrm{x}}} + \mathrm{c}$

• $\log \left({\mathrm{a}}^{\mathrm{x}} - 1\right) + \mathrm{c}$

If g(x) = $\frac{\mathrm{f}\left(\mathrm{x}\right) - \mathrm{f}\left(- \mathrm{x}\right)}{2}$  defined over [- 3, 3], and f(x) = 2x² - 4x + 1, then ${\int }_{- 3}^3\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}$ is equal to :

• 0

• 4

• - 4

• 8

$\int \frac{\sin \left(\mathrm{x}\right)}{\sin \left(\mathrm{x} - \mathrm{a}\right)}\mathrm{dx}$ is equal to :

• $\mathrm{xcos}\left(\mathrm{a}\right) - \sin \left(\mathrm{a}\right)\log \left(\sin \left(\mathrm{x} - \mathrm{a}\right)\right) + \mathrm{c}$

• $\mathrm{xsin}\left(\mathrm{a}\right) + \mathrm{c}$

• $\mathrm{xsin}\left(\mathrm{a}\right) + \sin \left(\mathrm{a}\right)\log \left(\sin \left(\mathrm{x} - \mathrm{a}\right)\right) + \mathrm{c}$

• $\mathrm{xcos}\left(\mathrm{a}\right) + \sin \left(\mathrm{a}\right)\log \left(\sin \left(\mathrm{x} - \mathrm{a}\right)\right) + \mathrm{c}$

$\int \frac{\mathrm{f}\text{'}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)\log \left[\mathrm{f}\left(\mathrm{x}\right)\right]}\mathrm{dx}$ is equal to :

• $\frac{\mathrm{f}\left(\mathrm{x}\right)}{\log \left(\mathrm{f}\left(\mathrm{x}\right)\right)}$

• f(x) . log(f(x)) + c

• $\log \left[\log \left(\mathrm{f}\left(\mathrm{x}\right)\right)\right] + \mathrm{c}$

• $\frac{1}{\log \left[\log \left(\mathrm{f}\left(\mathrm{x}\right)\right)\right]}$ + c

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

• $\log \left(\frac{1 + {\mathrm{e}}^{\mathrm{x}}}{2 + {\mathrm{e}}^{\mathrm{x}}}\right) + \mathrm{c}$

• $\log \left(\frac{2 + {\mathrm{e}}^{\mathrm{x}}}{1 + {\mathrm{e}}^{\mathrm{x}}}\right) + \mathrm{c}$

• $\left(\frac{1 + {\mathrm{e}}^{\mathrm{x}}}{2 + {\mathrm{e}}^{\mathrm{x}}}\right) + \mathrm{c}$

• $\left(\frac{2 + {\mathrm{e}}^{\mathrm{x}}}{1 + {\mathrm{e}}^{\mathrm{x}}}\right) + \mathrm{c}$

If I_n = ${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\tan }^{\mathrm{n}}\left(\mathrm{\theta }\right)\mathrm{d\theta }$, then ${\mathrm{I}}_{\mathrm{\theta }} + {\mathrm{I}}_6$ is equal to :

• $\frac{1}{7}$

• $\frac{1}{4}$

• $\frac{1}{5}$

• $\frac{1}{6}$

$\int \frac{\left({\mathrm{e}}^{\mathrm{x}} - {\mathrm{e}}^{- \mathrm{x}}\right)}{\left({\mathrm{e}}^{\mathrm{x}} + {\mathrm{e}}^{- \mathrm{x}}\right)\log \left(\cosh \left(\mathrm{x}\right)\right)}$ is equal to :

• $\log \left(\tanh \left(\mathrm{x}\right)\right)$

• $2\log \left({\mathrm{e}}^{\mathrm{x}} + {\mathrm{e}}^{- \mathrm{x}}\right) + \mathrm{c}$

• $2\log \left({\mathrm{e}}^{\mathrm{x}} - {\mathrm{e}}^{- \mathrm{x}}\right) + \mathrm{c}$

• $\mathrm{loglog}\left(\cosh \left(\mathrm{x}\right)\right) + \mathrm{c}$

${\int }_{- \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\cos \left(\mathrm{x}\right)\log \left(\frac{1 + \mathrm{x}}{1 - \mathrm{x}}\right)\mathrm{dx} = \mathrm{k} . \log \left(2\right)$, then k equals to

• 0

• - 1

• - 2

• $\frac{1}{2}$

${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{\cos \left(\mathrm{\theta }\right)}{\sqrt{4 - {\sin }^2\left(\mathrm{\theta }\right)}}\mathrm{d\theta }$ is equal to :

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{6}$

• $\frac{\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{5}$