Integration by Substitution: int (3x^2 +3)/(x^3 + 3x)^4 dx

[cmnalo]

∫ (3x^2 +3)/(x^3 + 3x)^4 dx

u= x^3 + 3x
du = 3x^2 +3dx
dx= du / (3x^2 +3)

∫ du / u^4

I'm confused as to my next step. The answer is -1/[3(x^3+3x)^3]+c

 

[skeeter]

\(\displaystyle \L \int u^{-4} du = -\frac{1}{3} u^{-3} + C\)

finish.