Evaluate $\int (2-3x^2)^{\frac{1}{3}}(-6x)dx$ using u-substitution

39 Views Asked by user637978 At 06 Apr 2026 - 5:31

Section 5.2

Let $u=2-3x^2$

Then $\frac{du}{dx}=-6x$ and so $\frac{du}{-6x}=dx$

Thus we have:

$\int (2-3x^2)^{\frac{1}{3}}(-6x)dx$

$= \int (u)^{\frac{1}{3}}(-6x)\frac{du}{-6x}$

$= \int (u)^{\frac{1}{3}}du$

$= \frac{u^{\frac{4}{3}}}{\frac{4}{3}} +C$

$= \frac{3}{4}(2-3x^2)^{\frac{4}{3}}+C$

Can somebody verify this for me?