<s>[INST] Integral of $\int{(x^2+2x)\over \sqrt{x^3+3x^2+1}} dx$ 
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!
 [/INST] You have $du=3(x^2+2x)dx$, so your integral become :
$$\int{(x^2+2x)\over \sqrt{x^3+3x^2+1}} dx=\int{1\over 3\sqrt{u}} du=\frac{1}{3}\int {1\over \sqrt{u}} du=\frac{2}{3}\sqrt u$$
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