Mathematics : Integrals

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

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{6}$

• $\frac{\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{5}$

If f(x) = $\cos \left(\mathrm{x}\right) - {\cos }^2\left(\mathrm{x}\right) + {\cos }^3\left(\mathrm{x}\right) - ... \mathrm{to} \infty$, then $\int \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$ equals to

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

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

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

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

${\int }_0^1\frac{\mathrm{xdx}}{\left[\mathrm{x} + \sqrt{1 - {\mathrm{x}}^2}\right]\sqrt{1 - {\mathrm{x}}^2}}$ is equal to :

• 0

• 1

• $\frac{\mathrm{\pi }}{4}$

• $\frac{{\mathrm{\pi }}^2}{2}$

The value of ${\int }_0^{2\mathrm{a}}\frac{\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}}{\mathrm{f}\left(\mathrm{x}\right) + \mathrm{f}\left(2\mathrm{a} - \mathrm{x}\right)}$ is :

• f(a)

• f(2a)

• f(0)

• a

The value of ${\int }_{- \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\mathrm{x}}^3{\sin }^4\left(\mathrm{x}\right)\mathrm{dx}$ is equal to :

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{8}$

• 0

For any positive integer n, $\int \frac{\mathrm{dx}}{{\mathrm{x}}^{\mathrm{n} + 1} + \mathrm{x}}$ is equal to :

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

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

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

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