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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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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 $u^2$.

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$$\begin{align} F(t) &= ∫(t^3*e^{(-t^2)})dt \\ u &= -t^2 \\ (-t^2)dt &= du \\ \left(\frac{d}{dt}(-t^2)\right)dt &= du \\ \left(-1*\frac{d}{dt}(t^2)\right)dt &= du \\ \left(-1*2t*\frac{d}{dt}(t)\right)dt &= du \\ \left(-1*2t*\frac{dt}{dt}\right)dt &= du \\ \left(-1*2t\right)dt &= du \\ \left(-2t\right)dt &= du \\ dt &= \left(\frac{1}{-2t}\right)du \\ dt &= \left(-\frac{1}{2t}\right)du \\ F(\ ) &= \int\left(t^3*e^u*-\frac{1}{2t}\right)du \\ &= -\frac{1}{2}*\int\left(t^3*e^u*\frac{1}{t}\right)du \\ &= -\frac{1}{2}*\int\left(t^2*e^u*\right)du \\ u &= -t^2 \\ -t^2 &= u \\ t^2 &= -u \\ F(u) &= -\frac{1}{2}*\int(-u*e^u)du \\ F(u) &= --\frac{1}{2}*\int(u*e^u)du \\ F(u) &= \frac{1}{2}*\int(u*e^u)du \\ \int wv' &= wv − \int w'v \\ w &= u \\ w' &= \frac{d}{du}(u) \\ w' &= \frac{du}{du} \\ w' &= 1 \\ v' &= e^u \\ v &= e^u \\ \int u*e^u &= u*e^u − \int1*e^u \\ \int u*e^u &= u*e^u − \int e^u \\ \int u*e^u &= e^u(u − 1) \\ F(u) &= \frac{1}{2}*e^u(u − 1) \\ F(t) &= \frac{1}{2}*e^{(-t^2)}(-t^2 − 1) \\ F(t) &= -\frac{1}{2}*e^{(-t^2)}(t^2 + 1) \\ F(t) &= \bf\left[-\frac{1}{2}*e^{(-t^2)}(t^2 + 1) + C \right]\\ \end{align}$$

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