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

Question

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

It is an improper integral and I know it diverges but I don't know why

 i found u=x^3-8

but am getting stuck after that

 

Answer 1

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.

Answer 2

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

Answer 3

∫4x^2/(x^3-8) dx

= ∫(4/3)/(x^3-8) d(x^3-8)

= (4/3)ln|x^3-8|, which approaches infinity as x --> 2.

So, the integral diverges.