# 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*}

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Try wolfram alpha. Click "show steps". –  Bill Cook Sep 19 '12 at 1:24

\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 $u^2$.