$\int {\mathrm{e}}^{3\log \left(\mathrm{x}\right)}{\left({\mathrm{x}}^4 + 1\right)}^{- 1}\mathrm{dx}$ is equal to

• e^3log(x) + c

• $\frac{1}{4}\log \left({\mathrm{x}}^4 + 1\right)$

• log(x⁴ + 1) + c

• $\frac{1}{2}\log \left({\mathrm{x}}^4 + 1\right) + \mathrm{c}$

If ${\int }_0^{\mathrm{\pi }}\mathrm{xf}\left(\sin \left(\mathrm{x}\right)\right)\mathrm{dx} = \mathrm{A}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{f}\left(\sin \left(\mathrm{x}\right)\right)\mathrm{dx}$, then A is

• 0

• $\mathrm{\pi }$

• $\frac{\mathrm{\pi }}{4}$

• 2$\mathrm{\pi }$

${\int }_{- 2}^2\left|\left[\mathrm{x}\right]\right|\mathrm{dx}$ is equal to

• 1

• 2

• 3

• 4

If $\int \frac{\sin \left(\mathrm{x}\right)}{\sin \left(\mathrm{x} - \mathrm{\alpha }\right)}\mathrm{dx} = \mathrm{Ax} + \mathrm{Blog}\left(\sin \left(\mathrm{x} - \mathrm{\alpha }\right)\right) + \mathrm{C}$, then value of A - B at $\mathrm{\alpha } = \frac{\mathrm{\pi }}{2}$ is

• - 1

• 1

• 2

• 0

If ${\int }_{\mathrm{a}}^{\mathrm{b}}{\mathrm{x}}^3\mathrm{dx} = 0 \mathrm{and} {\int }_{\mathrm{a}}^{\mathrm{b}}{\mathrm{x}}^2\mathrm{dx} = \frac{2}{3}$, then the values of a and b are respectively

• 1, - 1

• - 1, 1

• 1, 1

• - 1, - 1

$\int \left(\sqrt[3]{\mathrm{x}}\right)\left(\sqrt[5]{1 + \sqrt[3]{{\mathrm{x}}^4}}\right)\mathrm{dx}$ is equal to

• ${\left(1 + {\mathrm{x}}^{\frac{3}{4}}\right)}^{\frac{6}{5}} + \mathrm{c}$

• ${\left(1 + {\mathrm{x}}^{\frac{4}{3}}\right)}^{\frac{6}{5}} + \mathrm{c}$

• $\frac{5}{8}{\left(1 + {\mathrm{x}}^{\frac{4}{3}}\right)}^{\frac{6}{5}} + \mathrm{c}$

• $\frac{1}{6}{\left(1 + {\mathrm{x}}^{\frac{4}{3}}\right)}^6 + \mathrm{c}$

167.

If u = - $\mathrm{f}\text{'}\text{'}\left(\mathrm{\theta }\right)\sin \left(\mathrm{\theta }\right) + \mathrm{f}\text{'}\left(\mathrm{\theta }\right)\cos \left(\mathrm{\theta }\right)$ and v = $\mathrm{f}\text{'}\text{'}\left(\mathrm{\theta }\right)\cos \left(\mathrm{\theta }\right) + \mathrm{f}\text{'}\left(\mathrm{\theta }\right)\sin \left(\mathrm{\theta }\right)$, then $\int {\left[{\left(\frac{\mathrm{du}}{\mathrm{d\theta }}\right)}^2 + {\left(\frac{\mathrm{dv}}{\mathrm{d\theta }}\right)}^2\right]}^{\frac{1}{2}}\mathrm{d\theta }$ is equal to

• $\mathrm{f}\left(\mathrm{\theta }\right) - \mathrm{f}\text{'}\text{'}\left(\mathrm{\theta }\right) + \mathrm{c}$

• $\mathrm{f}\left(\mathrm{\theta }\right) + \mathrm{f}\text{'}\text{'}\left(\mathrm{\theta }\right) + \mathrm{c}$

• $\mathrm{f}\text{'}\left(\mathrm{\theta }\right) + \mathrm{f}\text{'}\text{'}\left(\mathrm{\theta }\right) + \mathrm{c}$

• $\mathrm{f}\text{'}\left(\mathrm{\theta }\right) - \mathrm{f}\text{'}\text{'}\left(\mathrm{\theta }\right) + \mathrm{c}$

$\int \frac{{\mathrm{e}}^{6{\log }_{\mathrm{e}}\left(\mathrm{x}\right)} - {\mathrm{e}}^{5{\log }_{\mathrm{e}}\left(\mathrm{x}\right)}}{{\mathrm{e}}^{4{\log }_{\mathrm{e}}\left(\mathrm{x}\right)} - {\mathrm{e}}^{3{\log }_{\mathrm{e}}\left(\mathrm{x}\right)}}\mathrm{dx}$ is equal to

• $\frac{{\mathrm{x}}^3}{3} + \mathrm{c}$

• $\frac{{\mathrm{x}}^2}{2} + \mathrm{c}$

• $\frac{{\mathrm{x}}^2}{3} + \mathrm{c}$

• $\frac{- {\mathrm{x}}^3}{3} + \mathrm{c}$

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

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

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

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

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

$\int \frac{{\mathrm{x}}^4 - 1}{{\mathrm{x}}^2{\left({\mathrm{x}}^4 + {\mathrm{x}}^2 + 1\right)}^{{\displaystyle \frac{1}{2}}}}\mathrm{dx}$ is equal to

• $\sqrt{\frac{{\mathrm{x}}^4 + {\mathrm{x}}^2 + 1}{\mathrm{x}}} + \mathrm{c}$

• $\frac{{\mathrm{x}}^2}{\sqrt{{\mathrm{x}}^4 + {\mathrm{x}}^2 + 1}} + \mathrm{c}$

• $\mathrm{x}{\left({\mathrm{x}}^4 + {\mathrm{x}}^2 + 1\right)}^{\frac{3}{2}} + \mathrm{c}$

• $\frac{\sqrt{{\mathrm{x}}^4 + {\mathrm{x}}^2 + 1}}{\mathrm{x}} + \mathrm{c}$