A derivative unlocks integration by parts

$\dfrac{d}{dx}(1-e^{4x})^{1/2}=\frac12(1-e^{4x})^{-1/2}(-4e^{4x})=-\dfrac{2e^{4x}}{\sqrt{1-e^{4x}}}$.

Integrate by parts with $u=\sin^{-1}(e^{2x})$ and $dv=e^{2x}dx$. Then $du=\dfrac{2e^{2x}}{\sqrt{1-e^{4x}}}dx$ and $v=\frac12e^{2x}$.

$\displaystyle\int e^{2x}\sin^{-1}(e^{2x})dx=\frac12e^{2x}\sin^{-1}(e^{2x})-\int\dfrac{e^{4x}}{\sqrt{1-e^{4x}}}dx$.

Part (i) gives $\displaystyle\int\dfrac{2e^{4x}}{\sqrt{1-e^{4x}}}dx=-\sqrt{1-e^{4x}}+C$, so the remaining integral is $-\frac12\sqrt{1-e^{4x}}$ plus a constant.

Therefore $\boxed{\frac12e^{2x}\sin^{-1}(e^{2x})+\frac12\sqrt{1-e^{4x}}+C}$.