antiderivative of (x^2+4x)^1/3 ?
\[\sqrt[3]{x^2+4x}\]
i know that u = x^2+4x so du = (2x+4)dx where do i go from there?
@agent0smith @Dido525
well u multipl that by 1/3 too
1/3 dx (x^2 + 4x) = 1/3* (2x + 4)..I think so
how did you get that? can u explain?
cuz du = (2x+4)dx = (2(x+2))dx so 1/2du = (x+2)dx then...?
actually, the value of power multplies..hmm du = 1/3*(2x + 4)dx => 3*du = (2x + 4)dx
I'm pretty sure you need to either do a Trig-Sub, or Integration-By-Parts might also be possible. No U-Sub available though :( Have you learned either of those methods yet?
yeah the substitution method.
Trig subs? :D
i think so? remind me what that is?
I think you might want to add and subtract +4 and -4 under the cube root so that you can factor.
google the concept behind it :)
idk how to do this, I'm just posting so that I get notified when someone answers it.
I could reduce it down so that it becomes \[2^{3/5}\int\limits_{}^{}\tan^{3/5} \theta \sec \theta d \theta\] after making the substitution: \[(x+2)=2\sec \theta\]
the fraction should be 5/3 not 3/5 on the 2 and tan(theta)
I can't seem to crack it. You're in calculus 1 or 2?
1
where did the tangent, secant stiff come from?
Ah sorry I was busy :C Kanoo been helping you? c:
I was doing trigonometric substitution, but you don't learn that until cal 2. If there's an easier way with just using u-sub then I would be surprised! Good luck zeppy! =P
Ya I dunno sorry Jenny :c This doesn't seem like a problem you should be doing yet, hmm.
Check this out: http://www.wolframalpha.com/input/?i=(x%5E2%2B4x)%5E1%2F3dx&t=crmtb01 I've never even heard of that stuff. Hypergeometric function? lol
Woah that looks fun :O
oy.....! -_-
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