## Calculus (3rd Edition)

$e^x(e^{2x}+1)^3- \frac{6}{7} e^{7x}- \frac{12}{5} e^{5x} -2e^{3x} +c$
We can do the integration by parts, $u=(e^{2x}+1)^3,v'=e^xdx$, then $u'=6e^{2x}(e^{2x}+1)^2dx$, $v= e^x$. Hence, we have $$\int e^x(e^{2x}+1)^3dx=e^x(e^{2x}+1)^3-6\int e^{3x}(e^{2x}+1)^2 dx\\\ =e^x(e^{2x}+1)^3-6\int e^{7x}+2e^{5x} +e^{3x} dx\\ =e^x(e^{2x}+1)^3- \frac{6}{7} e^{7x}- \frac{12}{5} e^{5x} -2e^{3x} +c .$$