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The integral in question looks like this: \begin{aligned} \large \ \int x^3 \cdot e^{8-7x^4} dx \end{aligned} I tried using u-substitution on all the part before dx and it eliminated e (with all its degrees) but left a huge mess: \begin{aligned} \ \int \frac {x^3}{3x^2-28x^6} du \end{aligned} I must have made a some simple mistake...

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    You know, this is almost identical to the other $q$uestion you just $a$sked and had answered. I'm sure that if you took the time to digest that other solution, you'd have no trou$b$le with this one.2012-02-19

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Remember that the chain rule says (e^{f(x)})'=f'(x)e^{f(x)}. In your case, $f(x)=8-7x^4$, f'(x)=-28x^3; from here you should be able to construct any anti-derivative explicitly.

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What is the derivative of $8−7x^4$?

Put $u = 8−7x^4$; then what is $du/dx$ ?

This should be sufficient starter for you.