C cmnalo Junior Member Joined Nov 5, 2006 Messages 61 Dec 4, 2006 #1 ∫ (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
∫ (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 Elite Member Joined Dec 15, 2005 Messages 3,204 Dec 4, 2006 #2 \(\displaystyle \L \int u^{-4} du = -\frac{1}{3} u^{-3} + C\) finish.
