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How do i integrate 4x^2/(x^3-8) from 0 to 8

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3 Answers

Or you can use a u substitution that you tried.

We have u = x3 - 8, and du = 3x2dx so

∫ [4x2/(x3 - 8)] dx = (4/3) ∫ du/u = (4/3) ln|u| + C = (4/3) ln|x3 - 8| + C

from 0 to 8 it diverges since at x = 2 you get (4/3) ln 0  → ∞ as said by Robert.

You can tell it diverges even before doing the integral, because at x = 2, the denominator of 4x2/(x3 - 8) is zero. There's a vertical asymptote at x = 2 (and that makes it difficult to find the area under the curve!).

Remember that that's the point of integration - to find the area under a curve, and you can't find the area under a vertical asymptote. Looking at a graph of the function may help to see this: 

https://www.google.com/search?q=4x%5E2%2F(x%5E3-8)&oq=4x%5E2%2F(x%5E3-8)&aqs=chrome.0.57&sourceid=chrome&ie=UTF-8