Evaluate $\int x e^{\sqrt{x}} \, dx$ $$\int_0^1 xe^{\sqrt{x}} dx = ? $$
All I can think of is the integration by parts rule, where 
$ u = x $ and $ dv= e^{\sqrt(x)} $ $ \Rightarrow du = 1$ and $ v= e^{\sqrt(x)} $ 
The answer I get is $e^{\sqrt(x)}(x-1)$ , which is wrong.
Can anyone please explain in detail?
 A: Let $u=\sqrt{x}$ so $du=\frac{1}{2\sqrt{x}}dx=\frac{1}{2u}dx$ so the integral becomes
$$ \int_0^1 2u^3e^u du$$
then evaluate this by integrating by parts.
A: You can do integration by parts like this, without substituting. Of course, substituting is fine and all, but you'll have to use Integration by Parts three times either way.
$$\begin{align}
\int_0^1xe^{\sqrt{x}}\,dx
&=\int_0^1\frac{2x\sqrt{x}}{2\sqrt{x}}e^{\sqrt{x}}\,dx\\
&=\int_0^12x\sqrt{x}\left(\frac{1}{2\sqrt{x}}e^{\sqrt{x}}\,dx\right)\\
&=\left[2x\sqrt{x}e^{\sqrt{x}}\right]_0^1-\int_0^13\sqrt{x}\,e^{\sqrt{x}}\,dx\\
&=\left[2x\sqrt{x}e^{\sqrt{x}}\right]_0^1-\int_0^13\sqrt{x}\,e^{\sqrt{x}}\,dx
\end{align}$$
And we've reduced the intebrand from $cx^1e^{\sqrt{x}}$ to $cx^{1/2}e^{\sqrt{x}}$. Repeat the technique two more times to bring it to $cx^{-1/2}e^{\sqrt{x}}$, and then you can just directly antidifferentiate.
A: In fact,
$$
\int xe^{\sqrt{x}}dx = 2\int x\sqrt{x}\dfrac{e^{\sqrt{x}}}{2\sqrt{x}}dx = 2\int x\sqrt{x} e^{\sqrt{x}}d(\sqrt{x}) = 2\int u^{3}e^u du = \dfrac{2}{D}(u^3e^u)
$$
$$
= 2e^u \dfrac{1}{1 + D}u^3 = 2e^u(1 - D + D^2 - D^3)u^3 = 2e^u[u^3 - 3u^2 + 6u - 6]
$$
$$
=2e^{\sqrt{x}}[x^{3/2} - 3x + 6\sqrt{x} - 6] + C
$$
