# Integrating $\int\frac{5x^2}{(2+3x^3)^2}\ dx$ [closed]

I would like to know how I should calculate the integration below.

How do you integrate it?

$$\int\frac{5x^2}{(2+3x^3)^2}\ \mathrm{d}x$$

## closed as off-topic by Did, kingW3, user8795, Shaun, projectilemotionFeb 7 '17 at 19:43

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• Please share your thoughts so far :) – Shaun Feb 7 '17 at 18:22

Substitute $u = 2+3x^3$, so that $du = 9x^2 dx$.
Then, $\displaystyle \frac{5}{9}\int \frac{9x^2}{u^2} \,dx = \frac{5}{9}\int \frac{1}{u^2} \, du = \left(\frac{5}{9}\right)\left(-\frac{1}{u}\right) = \boxed{-\frac{5}{9(2+3x^3)}+C}$.
Assume $2+3x^3=t$ which gives $9x^2dx=dt$ or $x^2dx=dt/9$ $$\int\frac{5x^2}{(2+3x^3)^2}dx=\int\frac{5/9}{t^2}dt=-\frac{5}{9t}+C=-\frac{5}{9(2+3x^3)}+C$$
Another way instead of using substitution. $$\int\frac{5x^2}{(2+3x^3)^2}\, \mathrm{d}x=\frac{5}{3}\int\frac{1}{(2+3x^3)^2}\, \mathrm{d}\left ( x^{3} \right )=\frac{5}{9}\int\frac{1}{(2+3x^3)^2}\, \mathrm{d}\left ( 2+3x^{3} \right )=-\frac{5}{9\left ( 2+3x^{3} \right )}+C$$