Question

Integrate the function: $\frac{x^2}{(2+3x^3)^3}$

Answer 1

Let 2 + 3x3 = t

∴ 9x2 dx = dt

$\Rightarrow \frac{x^2}{(2+3x^3)}dx=\frac{1}{9}\int\frac{dt}{(t)^3}$

= $\frac{1}{9}\bigg[\frac{t^{-2}}{-2}\bigg]+C$

= $\frac{-1}{18}\bigg(\frac{1}{t^2}\bigg)+C$

=$\frac{-1}{18(2+3x^3)^2}+C$