Mathematics : Integrals

If ${\int }_0^{\infty }\frac{\log \left(1 + {\mathrm{x}}^2\right)}{1 + {\mathrm{x}}^2}d\mathrm{x} = \mathrm{k}{\int }_0^1\frac{\log \left(1 + \mathrm{x}\right)}{1 + {\mathrm{x}}^2}\mathrm{dx}$, then k is equal to

• 4

• 8

• $\mathrm{\pi }$

• 2$\mathrm{\pi }$

${\int }_0^{\mathrm{x}}\left|\sin \left(\mathrm{t}\right)\right|\mathrm{dt}, \mathrm{where} \mathrm{x} \in \left[2\mathrm{n\pi }, \left(2\mathrm{n} + 1\right)\mathrm{\pi }\right], \mathrm{n} \in \mathrm{N}$, is equal to

• 4n - 1 - cos(x)

• 4n - sin(x)

• 4n - cos(x)

• 4n + 1 - cos(x)

Let I₁ = ${\int }_0^1\frac{{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}}{1 + \mathrm{x}}$ and I₂ = ${\int }_0^1\frac{{\mathrm{x}}^2\mathrm{dx}}{{\mathrm{e}}^{{\mathrm{x}}^3}\left(2 - {\mathrm{x}}^3\right)}$ Then, $\frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}$ is equal to

• $\frac{1}{3\mathrm{e}}$

• 3e

• $\frac{\mathrm{e}}{3}$

• $\frac{3}{\mathrm{e}}$

${\int }_{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^5\frac{\sqrt{{\left(25 - {\mathrm{x}}^2\right)}^3}}{{\mathrm{x}}^4}\mathrm{dx}$ is equal to

• $\frac{\mathrm{\pi }}{3}$

• $\frac{2\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{6}$

• $\frac{5\mathrm{\pi }}{6}$

$\int \frac{\mathrm{dx}}{\cos \left(\mathrm{x}\right) + \sqrt{3}\sin \left(\mathrm{x}\right)}$ equals

• $\frac{1}{2}\log \left(\tan \left(\frac{\mathrm{x}}{2} + \frac{\mathrm{\pi }}{12}\right)\right) + \mathrm{C}$

• $\frac{1}{3}\log \left(\tan \left(\frac{\mathrm{x}}{2} - \frac{\mathrm{\pi }}{12}\right)\right) + \mathrm{C}$

• $\log \left(\tan \left(\frac{\mathrm{x}}{2} + \frac{\mathrm{\pi }}{6}\right)\right) + \mathrm{C}$

• $\frac{1}{2}\log \left(\tan \left(\frac{\mathrm{x}}{2} - \frac{\mathrm{\pi }}{6}\right)\right) + \mathrm{C}$

${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sin \left(2\mathrm{x}\right) . \log \left(\tan \left(\mathrm{x}\right)\right)\mathrm{dx}$

• 0

• 2

• 4

• 7

If $∆\mathrm{x} = \left|\begin{array}{ccc}1& \cos \left(\mathrm{x}\right)& 1 - \cos \left(\mathrm{x}\right)\\ 1 + \sin \left(\mathrm{x}\right)& \cos \left(\mathrm{x}\right)& 1 + \sin \left(\mathrm{x}\right) + \cos \left(\mathrm{x}\right)\\ \sin \left(\mathrm{x}\right)& \sin \left(\mathrm{x}\right)& 1\end{array}\right|$, then ${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}∆\left(\mathrm{x}\right)\mathrm{dx}$ is equal to

• $\frac{1}{4}$

• $\frac{1}{2}$

• 0

• $- \frac{1}{4}$

${\int }_0^{2\mathrm{\pi }}\left(\sin \left(\mathrm{x}\right) + \left|\sin \left(\mathrm{x}\right)\right|\right)\mathrm{dx}$ is equal to

• 0

• 4

• 8

• 1

The value of ${\int }_0^{\sqrt{2}}\left[{\mathrm{x}}^2\right]\mathrm{dx}$, where [.] is the greatest integer function, is

• 2 - $\sqrt{2}$

• 2 + $\sqrt{2}$

• $\sqrt{2}$ - 1

• $\sqrt{2}$ - 2

If l (m,n) = ${\int }_0^1{\mathrm{t}}^{\mathrm{m}}{\left(1 + \mathrm{t}\right)}^{\mathrm{n}}\mathrm{dt}$, then the expression for l (m, n) in terms of l (m + 1, n + 1) is

• $\frac{2^{\mathrm{n}}}{\mathrm{m} + 1} - \frac{\mathrm{n}}{\mathrm{m} + 1} . \mathrm{l} \left(\mathrm{m} + 1, \mathrm{n} - 1\right)$

• $\frac{\mathrm{n}}{\mathrm{m} + 1} . \mathrm{l} \left(\mathrm{m} + 1, \mathrm{n} - 1\right)$

• $\frac{2\mathrm{n}}{\mathrm{m} + 1} + \frac{\mathrm{n}}{\mathrm{m} + 1} . \mathrm{l} \left(\mathrm{m} + 1, \mathrm{n} - 1\right)$

• $\frac{\mathrm{m}}{\mathrm{n} + 1} . \mathrm{l} \left(\mathrm{m} + 1, \mathrm{n} - 1\right)$