\[\int \frac{x^3-2}{x^3+2} \cdot x^2 dx\]

The trick to this one is to rewrite the numerator as \(x^3+2-4\) then separate the function into two integrals, this is because \(-2=+2-4\).

Then set \(u=x^3+2\) and \(du=3x^2 dx\), then multiply both sides by \(\frac{4}{3}\) for du.

Challenging Integrals

\[\int sec(x) dx\]

For more challenging integrals, look up “Integration Bee”.