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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 should retry the u substitution with $u = 8 - 7x^4$. What is $du$ in this case? The answer you get should be in the form $C \int e^u du$ for some constant $C$. Recall from the last thread that the idea is to look for a function and its derivative. – Chris Janjigian Feb 19 '12 at 23:19
You know, this is almost identical to the other question you just asked and had answered. I'm sure that if you took the time to digest that other solution, you'd have no trouble with this one. – Gerry Myerson Feb 19 '12 at 23:21

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.

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