$\int \frac{e^x \, dx}{\sqrt{1 - e^{2x}}} = $

$\int {\frac{{{e^{{{\tan }^{ - 1}}\sqrt x }}}}{{\sqrt x + x\sqrt x }}} dx = $

If $x \neq -1$ and $\int \frac{x^3+x^2-x-1}{(x^5+x^4+3x^3+3x^2+x+1) \tan^{-1}(\frac{x^2+1}{x})} dx = A \log(f(x)) + C$, then $A - \tan(f(2)) = $

If $f(x) = \frac{x}{(1 + nx^n)^{1/n}}$ for $n \geq 2$,then $\int x^{n-2} f(x) dx =$

Let the equation of the curve passing through the point $(0,1)$ be given by $y=\int x^3 e^{x^4} d x$. If the equation of the curve is written in the form $x=f(y)$,then $f(y)=$

$\int {\frac{{\sqrt {{x^2} + 1} [\log ({x^2} + 1) - 2\log x]}}{{{x^4}}}} dx$ is equal to