$\int_{0}^{\infty} e^{-x} \sin^{6} x dx =$

If $\int(\sin x )^{\frac{-11}{2}}(\cos x )^{\frac{-5}{2}} dx = -\frac{p_1}{q_1}(\cot x)^{\frac{9}{2}}-\frac{p_2}{q_2}(\cot x)^{\frac{5}{2}}-\frac{p_3}{q_3}(\cot x)^{\frac{1}{2}}+\frac{p_4}{q_4}(\cot x)^{\frac{-3}{2}}+C,$ where $p_i$ and $q_i$ are positive integers with $\operatorname{gcd}(p_i, q_i)=1$ for $i =1,2,3,4$ and $C$ is the constant of integration, then $\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}$ is equal to . . . . . . .

$\int \sqrt{x^2+x+1} \, dx \times \int \frac{1}{\sqrt{x^2+x+1}} \, dx$ is equal to

$\int \left[ \log(\log x) + \frac{1}{(\log x)^2} \right] dx = $

For real numbers $\alpha, \beta, \gamma$ and $\delta,$ if $\int \frac{\left(x^{2}-1\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)}{\left(x^{4}+3 x^{2}+1\right) \tan ^{-1}\left(\frac{x^{2}+1}{x}\right)} d x =\alpha \log _{e}\left(\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right) +\beta \tan ^{-1}\left(\frac{\gamma\left(x^{2}-1\right)}{x}\right)+\delta \tan ^{-1}\left(\frac{x^{2}+1}{x}\right)+C$ where $C$ is an arbitrary constant,then the value of $10(\alpha+\beta \gamma+\delta)$ is equal to ....... .

If $\int(1+x) \log \left(1+x^2\right) d x=\left(x+\frac{x^2}{2}+\frac{1}{2}\right) \log \left(1+x^2\right)+g(x)+C$,then $g(x)=$