Indefinite integral $\int{3x^2\over (x^3+2)^4}dx$

Question

Question:

How to solve this indefinite interal $$\int{3x^2\over (x^3+2)^4}dx$$

Attempt:

$3\int x^2{1\over (x^3+2)^4}dx$

Let $u=x^2$ then $du={1\over (x^3+2)^4}dx$.

Am I on the right track or going about it the wrong way?

ok so I make $du=3x^2dx$ but what do I do with this? Does it just disappear?

Answer 1

Hint: Let $u = x^3 + 2$ then $du = 3x^2 dx$ and your integral becomes

$$\int \frac{1}{u^4} du$$

Answer 2

Your option $u=x^2$ seems to be a bad one. Indeed, $du=2x\,dx$ doesn't mate very well with the given numerator, and the denominator will turn to the unappetizing $(\sqrt u^3+2)^4$.

It is much more appealing to notice that $3x^2$ is the derivative of $x^3$, so that a substitution $u=x^3$ will yield $du$ at the numerator and $(u+2)^4$ at the denominator, a much more tractable result.

Answer 3

The substitution $u = x^3 +2$ and $\mathrm{d}u = 3x^2 \, \mathrm{d}x$ giving $$\int \frac{1}{u^4} \, \mathrm{d}u = -\frac{1}{3u^3} = \bbox[10px, border: solid blue 1px]{-\frac{1}{3(x^3+2)^3} + \mathrm{C}}$$

is much more elegant and tractable than the substitution you proposed.