Answer
$$\int \frac{1}{\sqrt xe^{-\sqrt x}}\sec^2(e^{\sqrt x}+1)dx=2\tan(e^{\sqrt x}+1)+C$$
Work Step by Step
$$A=\int \frac{1}{\sqrt xe^{-\sqrt x}}\sec^2(e^{\sqrt x}+1)dx$$ $$A=\int\frac{e^{\sqrt x}}{\sqrt x}\sec^2(e^{\sqrt x}+1)dx$$
We set $u=e^{\sqrt x}+1$
Then $$du=e^{\sqrt x}(\sqrt x)'dx=\frac{e^{\sqrt x}}{2\sqrt x}dx$$ $$\frac{e^{\sqrt x}}{\sqrt x}dx=2du$$
Therefore, $$A=2\int\sec^2udu=2\tan u+C $$ $$A=2\tan(e^{\sqrt x}+1)+C$$
