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**Evaluate the following definite integral.**

\(\displaystyle \int^1_0 36x^2(x^3 + 1)^4\)

\(\displaystyle u = x^3 + 1\)

\(\displaystyle du = 3x^2\)

But I don't have a \(\displaystyle 3x^2\) up top, I have a \(\displaystyle 36x^2\) so I divided out and got

\(\displaystyle \frac{du}{3} = x^2\)

now I update the limits and I get

\(\displaystyle 0^3 + 1 = 1\)

\(\displaystyle 1^3 + 1 = 2\)

so

\(\displaystyle \frac{1}{3} \int^2_1 du * u^4 = \frac{1}{3} * \frac{1}{5}u^5 = \frac{1}{15} * (2^3 + 1)^4] - [\frac{1}{15} * (1^3 + 1)^4]\)

Does this look correct? or have i messed up somewhere? I mean really i have to do \(\displaystyle 9^4\)? (sorry if that sounds ignorant, just want to make sure)