I'm having trouble evaluating this indefinite integral; even when I followed it through step by step, the answer I obtain is not correct. I typed it up on this text, it's kind of hard to read (I don't have any experience with LaTeX) but I tried to make it as clear as possible; I'd really appreciate if someone could point out my mistake(s):
$$\int t^3 e^{-t^2}dt$$
Let $u=t^2$; $du=2t~dt$, so $dt=\dfrac{du}{2t}=\dfrac{du}{2u^{1/2}}$. Then
$$\begin{align*}
\int t^3 e^{-t^2}dt&=\int u^{3/2} e^{-u} \frac{du}{2u^{1/2}}\\
&=\int u^2 e^{-u}\frac{du}2\\
&=\frac12\int u^2 e^{-u}du\;.
\end{align*}$$
Now integrate by parts:
$$\begin{array}{cc}
a=u^2&db=e^{-u}du\\
da=2u~du&b=-e^{-u}
\end{array}$$
$$\begin{align*}
ab-\int b~da&=u^2\left(-e^{-u}\right)-\int\left(-e^{-u}\right)(2u)du\\
&=u^2\left(-e^{-u}\right)+\int e^{-u}(2u)du\\
&=u^2\left(-e^{-u}\right)+2\int e^{-u}u~du\;.
\end{align*}$$
Another integration by parts:
$$\begin{array}{cc}
a=u&db=e^{-u}du\\
da=du&b=-e^{-u}
\end{array}$$
$$\begin{align*}
ab-\int b~da&=uu\left(-e^{-u}\right)-\int\left(-e^{-u}\right)du\\
&=u\left(-e^{-u}\right)+\int e^{-u}du\\
&=u\left(-e^{-u}\right)+2\left[u\left(-e^{-u}\right)\right]+C\;.
\end{align*}$$
Best Answer
You have an algebra error here:
$$\begin{align*} \int t^3 e^{-t^2}dt&=\int u^{3/2} e^{-u} \frac{du}{2u^{1/2}}\\ &=\int u^2 e^{-u}\frac{du}2\;: \end{align*}$$
$\dfrac{u^{3/2}}{u^{1/2}}=u$, not $u^2$.