<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] You could have made life simpler getting rid of the radical
$$u=\sqrt{7+e^x}\implies x=\log \left(u^2-7\right)\implies dx=\frac{2 u}{u^2-7}\,du$$
$$e^{2x}\sqrt{7+e^x}\,dx=2u^2(u^2-7)\,du$$
$$\int e^{2x}\sqrt{7+e^x}\,dx=2\int (u^4-7u^2)\,du=\frac25u^5-\frac 73 u^3+ C$$
Back to $x$
$$\int e^{2x}\sqrt{7+e^x}\,dx=\frac{2}{15} \sqrt{e^x+7} \left(3 e^{2 x}+7 e^x-98\right)+C$$
</s>