I get a final answer of $\frac{3\ln \left | e^x -3 \right |}{10} - \frac{3\ln \left | e^{2x} +1 \right |}{20} +\frac{\ln \left | e^{2x} + 1 \right |}{20} + C$ but it seems off, any help would be appreciated!
I used rationalizing substitution then partial fractions.
My partial fractions go as follows: $\frac{u}{(u-3)(u^2 +1)}= \frac{3}{10(u-3)} + \frac{-3u+1}{10(u^2 +1)}$
\begin{align} \int\frac{dx \,e^x}{(e^x-3)(e^{x^2}+1)} &= \int\frac{du}{(u-3)(u^2+1)}\tag{$u=e^x$} \\ &= \int du \, \frac{-3u+1}{10(u^2+1)}+\frac{3}{10(u-3)} \\ &= \int du \, \frac{-3u}{10(u^2+1)}+\int du \,\frac{1}{10(u^2+1)}+\int du \,\frac{3}{10(u-3)} \\ &= -\frac {3}{20}\log|u^2+1|+\frac{\tan^{-1} u}{10}+ \frac{3\log |u-3|}{10}+C \\ &= -\frac {3}{20}\log|e^{2x}+1|+\frac{\tan^{-1} e^x}{10}+ \frac{3\log |e^x-3|}{10}+C \end{align}