Proper integration procedure Is this the proper procedure for the below integral?
$$\int xe^{(x^2)}dx$$
let $u=e^x$
$du=xdx$
$$=\int u^2du$$
$$={u^3\over 3}+c$$
$$={(e^x)^3\over 3}+c$$
Wolfram alpha gives me this answer
$={(e^x)^2\over 2}+c$
 A: The substitution used in the question is not proper. If we use the substitution $u=e^x$, then $\mathrm{d}u=e^x\,\mathrm{d}x$, not $x\,\mathrm{d}x$.
Integration by parts with $u=x$ and $v=\frac12e^{2x}$ gives
$$
\begin{align}
\int x(e^x)^2\,\mathrm{d}x
&=\int xe^{2x}\,\mathrm{d}x\\
&=\frac12\int x\,\mathrm{d}e^{2x}\\
&=\frac12xe^{2x}-\frac12\int e^{2x}\,\mathrm{d}x\\
&=\frac12xe^{2x}-\frac14e^{2x}+C
\end{align}
$$

If, on the other hand, you were trying to integrate $xe^{x^2}$, then we can use $u=x^2$
$$
\begin{align}
\int xe^{x^2}\,\mathrm{d}x
&=\frac12\int e^u\,\mathrm{d}u\\
&=\frac12 e^u+C\\
&=\frac12 e^{x^2}+C
\end{align}
$$
A: Assuming you want to integrate $x(e^2x)$, then a simple wack with the integration by parts hammer, with $u = x$ and $\mathrm{d}v = e^{2x}$ so that you get $$\begin{align}\int xe^{2x} \, \mathrm{d}x &= \frac{1}{2}xe^{2x} - \frac{1}{2}\int e^{2x} \, \mathrm{d}x \\ &= \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} + \mathrm{C} \\ \\ & \bbox[border: blue 1px solid, 10px]{= \frac{e^{2x}}{4}\left(2x - 1\right) + \mathrm{C}} \end{align}$$

On the other hand, if you wanted to integrate $xe^{x^2}$, then you'd make the substitution $$u= x^2 \implies \mathrm{d}u = 2x \, \mathrm{d}x$$ which makes your integral much more tractable. Your transformed integral is then $$\int \frac{1}{2}e^u \, \mathrm{d}u = \frac{1}{2}e^{u} + \mathrm{C}$$
Back substitution yields $$\bbox[border: 1px blue solid, 10px]{\int xe^{x^2}\, \mathrm{d}x = \frac{1}{2}e^{x^2} + \mathrm{C}}$$
A: First, you really need to pay attention on how to set the brackets. As you can see, the answers are changing completely. Your procedure - substitution - is a good choice, but you made a mistake while substituting. Your $du=x dx$ is wrong it should be $du=e^x dx$, but still, this won't help you much, you're better off with $u=x^2$, just look below.
Assuming the OP wants to calculate
$$
\int x(e^x)^2dx=\int xe^{2x}dx
$$
then you probably want to use partial integration - also called integration by parts, which states - the $(')$ indicates the derivative -
$$
\int u v^{'}dx=uv-\int v u^{'}dx
$$
now we set $u=x$ and $v^{'}=e^{2x}$ which leads to $u^{'}=1$ and $v=\frac 1 2e^{2x}$ and therefore we get 
$$
\int u v^{'}dx=\int xe^{2x}dx=\frac 1 2xe^{2x}-\int\frac 1 2 e^{2x}dx=\frac 1 2xe^{2x}-\frac 1 4e^{2x}+c=\frac 1 2e^{2x}(x-\frac 1 2)+c
$$
EDIT if you want to calculate $\int x e^{x^2}dx$
Just for the sake of completeness, now the integral with substitution $u={x^2}, du=2xdx$
$$
\int x e^{x^2}dx=\int x e^{u}\frac 1 {2x}du=\frac 1 2e^u+c=\frac1 2e^{x^2}+c
$$
