<s>[INST] Evaluate $\int e^{2x}(7+e^x)^{1/2}\,dx$ $\int e^{2x}(7+e^x)^{1/2}\,dx$
Let $u=7+e^x \rightarrow du=e^xdx$
So the integral becomes:
$\int u^{\frac{3}{2}}-7u^{\frac{1}{2}}du$ and so our answer is
$\frac{2}{5}(7+e^x)^{\frac{5}{2}}-\frac{14}{3}(7+e^x)^{\frac{3}{2}}+C$
But this is not what wolfram says. Did I make a mistake?
 [/INST] The integral becomes
\begin{align}
\int (u-7)u^{\frac{1}{2}}du
&=\frac{2}{5}(7+e^{x})^{\frac{5}{2}}-\frac{14}{3}(7+e^{x})^{\frac{3}{2}}+C\\
&=\frac{2}{15}(7+e^{x})^{\frac{1}{2}}(3(7+e^{x})^{2}-35(7+e^{x}))+C\\
&=\frac{2}{15}\sqrt{7+e^{x}}(7e^{x}+3e^{2x}-98)+C
\end{align}
as required.
</s>