Integration with substitution and by parts 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:
$$\begin{array}{cc}
a=u&db=e^{-u}du\\
da=du&b=-e^{-u}
\end{array}$$
$$\begin{align*}
ab-\int b~da&=uu\left(-e^{-u}\right)-\int\left(-e^{-u}\right)du\\
&=u\left(-e^{-u}\right)+\int e^{-u}du\\
&=u\left(-e^{-u}\right)+2\left[u\left(-e^{-u}\right)\right]+C\;.
\end{align*}$$
 A: 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 $u^2$.
A: $$\begin{align}
F(t) &= ∫(t^3*e^{(-t^2)})dt \\
  u &= -t^2
 \\
  (-t^2)dt &= du \\
  \left(\frac{d}{dt}(-t^2)\right)dt &= du \\
  \left(-1*\frac{d}{dt}(t^2)\right)dt &= du \\
  \left(-1*2t*\frac{d}{dt}(t)\right)dt &= du \\
  \left(-1*2t*\frac{dt}{dt}\right)dt &= du \\
  \left(-1*2t\right)dt &= du \\
  \left(-2t\right)dt &= du \\
  dt &= \left(\frac{1}{-2t}\right)du \\
  dt &= \left(-\frac{1}{2t}\right)du \\
F(\ ) &= \int\left(t^3*e^u*-\frac{1}{2t}\right)du \\
      &= -\frac{1}{2}*\int\left(t^3*e^u*\frac{1}{t}\right)du \\
      &= -\frac{1}{2}*\int\left(t^2*e^u*\right)du \\
  u &= -t^2 \\
  -t^2 &= u \\
  t^2 &= -u \\
F(u) &= -\frac{1}{2}*\int(-u*e^u)du \\
F(u) &= --\frac{1}{2}*\int(u*e^u)du \\
F(u) &= \frac{1}{2}*\int(u*e^u)du \\
  \int wv' &= wv − \int w'v \\
  w &= u \\
  w' &= \frac{d}{du}(u) \\
  w' &= \frac{du}{du} \\
  w' &= 1 \\
  v' &= e^u \\
  v &= e^u \\
  \int u*e^u &= u*e^u − \int1*e^u \\
  \int u*e^u &= u*e^u − \int e^u \\
  \int u*e^u &= e^u(u − 1) \\
F(u) &= \frac{1}{2}*e^u(u − 1) \\
F(t) &= \frac{1}{2}*e^{(-t^2)}(-t^2 − 1) \\
F(t) &= -\frac{1}{2}*e^{(-t^2)}(t^2 + 1) \\
F(t) &= \bf\left[-\frac{1}{2}*e^{(-t^2)}(t^2 + 1) + C \right]\\
\end{align}$$
