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Evaluate $$\! \int t^2e^{-t^3} \ \mathrm{d}t$$ using integration by parts.

I know how to use substitution to evaluate this integral, when I use integration by parts letting $u = e^{-t^3}, \mathrm{d}v = t^2\ \mathrm{d}t$ I get $$\frac13t^3e^{-t^3}+\int\! t^5e^{-t^3}\ \mathrm{d}t$$ but this leads me to nowhere. Thanks

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  • $\begingroup$ Integration by parts is not necessary to evaluate this integral. But the usual strategy is to choose a different $u$, $\mathsf dv$ if the first choice didn't work. Here is a resource that may prove useful in determining how to choose $u$ and $\mathsf dv$ mathnow.wordpress.com/2009/10/14/liate-ilate-and-detail $\endgroup$
    – Math1000
    Sep 10, 2015 at 3:17

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We have $$ \int_{t} t^{2}e^{-t^{3}} = \int_{t} De^{-t^{3}}(\frac{-1}{3}) = \frac{-1}{3}e^{-t^{3}} + C $$ by the chain rule and fundamental theorem of calculus.

I am not sure why you are required to do it by integration by parts. But, in this case, there is no need to use it...

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