How do I evaluate the following integral?

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Answer 1 · 2018-02-26T20:24:18

#1/2x^2+18ln|x^2-36|+C#

Divide #x^2-36# into #x^3# using polynomial long division.

This yields:

#intx^3/(x^2-36)dx=int(x+(36x)/(x^2-36))dx#

Split this integral up:

#intxdx + int(36x)/(x^2-36)dx#

#intxdx=1/2x^2#

For #int(36x)/(x^2-36)dx#

#u=x^2-36#

#du=2xdx#

#18du=36xdx#

So, our integral becomes:

#1/2x^2+18int(du)/u=1/2x^2+18ln|u|+C=1/2x^2+18ln|x^2-36|+C#