Answer
$2\sqrt{x}e^{\sqrt{x}}-2e^{\sqrt{x}}+c$
Work Step by Step
Substitute $a=\sqrt{x}$.
If $a=\sqrt{x}$, then $da=\frac{1}{2\sqrt{x}}dx$.
Solve for $dx$ to get $dx=2\sqrt{x}\thinspace da$. Since $\sqrt{x}=a$, $dx=2a\thinspace da$
$\int e^{\sqrt{x}}dx=\int e^a2a\thinspace da$
$=2\int ae^a\thinspace da$
Now we'll use integration by parts. Choose $u=a$ and $dv=e^a\thinspace da$.
$u=a$
$du=da$
$dv=e^a\thinspace da$
$v=e^a$
So 2$\int ae^a\thinspace da=2(ae^a-\int e^a\thinspace da)$
$=2ae^a-2e^a$
Substituting $a=\sqrt{x}$ back in, we get $2\sqrt{x}e^{\sqrt{x}}-2e^{\sqrt{x}}$.
