Integral of $\int{(x^2+2x)\over \sqrt{x^3+3x^2+1}} dx$

Question

Find the integral of the following:

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

Do set $u=x^3+3x^2+1$?

So, $du=(3x^2+6x)dx$?

And, $x^2+2x={u-1-x^2\over x}$?

So then,

$$\int{({u-1-x^2\over x})\over \sqrt u} du$$

This seems very complicated, is there any easier way to do this? or if this is correct, could anyone show me how to go on with this? Thanks!

Answer 1

$du = (3x^2+6x)dx = 3(x^2+2x)dx$, so your answer changes as $\frac{1}{3}\int \frac{du}{\sqrt{u}} = \frac{2}{3}\sqrt{u}$

Answer 2

You have $du=3(x^2+2x)dx$, so your integral become :

$$\int{(x^2+2x)\over \sqrt{x^3+3x^2+1}} dx=\int{1\over 3\sqrt{u}} du=\frac{1}{3}\int {1\over \sqrt{u}} du=\frac{2}{3}\sqrt u$$