0
$\begingroup$

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

How do you integrate it?

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

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, kingW3, user8795, Shaun, projectilemotion
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Please share your thoughts so far :) $\endgroup$ – Shaun Feb 7 '17 at 18:22
11
$\begingroup$

Substitute $u = 2+3x^3$, so that $du = 9x^2 dx$.

Then, $\displaystyle \frac{5}{9}\int \frac{9x^2}{u^2} \,dx = \frac{5}{9}\int \frac{1}{u^2} \, du = \left(\frac{5}{9}\right)\left(-\frac{1}{u}\right) = \boxed{-\frac{5}{9(2+3x^3)}+C}$.

$\endgroup$
3
$\begingroup$

Assume $2+3x^3=t$ which gives $9x^2dx=dt$ or $x^2dx=dt/9$ $$\int\frac{5x^2}{(2+3x^3)^2}dx=\int\frac{5/9}{t^2}dt=-\frac{5}{9t}+C=-\frac{5}{9(2+3x^3)}+C$$

$\endgroup$
2
$\begingroup$

Another way instead of using substitution. $$\int\frac{5x^2}{(2+3x^3)^2}\, \mathrm{d}x=\frac{5}{3}\int\frac{1}{(2+3x^3)^2}\, \mathrm{d}\left ( x^{3} \right )=\frac{5}{9}\int\frac{1}{(2+3x^3)^2}\, \mathrm{d}\left ( 2+3x^{3} \right )=-\frac{5}{9\left ( 2+3x^{3} \right )}+C$$

$\endgroup$
  • $\begingroup$ This is "using substitution", only with different notations. $\endgroup$ – Did Jan 12 '17 at 18:22

Not the answer you're looking for? Browse other questions tagged or ask your own question.