antiderivative of (x^2+4x)^1/3 ?

- jennychan12

antiderivative of (x^2+4x)^1/3 ?

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- jennychan12

\[\sqrt[3]{x^2+4x}\]

- jennychan12

i know that u = x^2+4x
so du = (2x+4)dx
where do i go from there?

- jennychan12

@agent0smith @Dido525

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## More answers

- anonymous

well u multipl that by 1/3 too

- anonymous

1/3 dx (x^2 + 4x) = 1/3* (2x + 4)..I think so

- jennychan12

how did you get that? can u explain?

- jennychan12

cuz du = (2x+4)dx = (2(x+2))dx
so 1/2du = (x+2)dx
then...?

- anonymous

actually, the value of power multplies..hmm
du = 1/3*(2x + 4)dx => 3*du = (2x + 4)dx

- zepdrix

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?

- jennychan12

yeah the substitution method.

- zepdrix

Trig subs? :D

- jennychan12

i think so? remind me what that is?

- Kainui

I think you might want to add and subtract +4 and -4 under the cube root so that you can factor.

- anonymous

google the concept behind it :)

- beginnersmind

idk how to do this, I'm just posting so that I get notified when someone answers it.

- Kainui

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:
\[(x+2)=2\sec \theta\]

- Kainui

the fraction should be 5/3 not 3/5 on the 2 and tan(theta)

- anonymous

this link will help you :)
http://www.the-mathroom.ca/freebs/cali3/cali3.htm

- Kainui

I can't seem to crack it. You're in calculus 1 or 2?

- jennychan12

1

- jennychan12

where did the tangent, secant stiff come from?

- zepdrix

Ah sorry I was busy :C Kanoo been helping you? c:

- Kainui

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

- zepdrix

Ya I dunno sorry Jenny :c
This doesn't seem like a problem you should be doing yet, hmm.

- Kainui

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

- zepdrix

Woah that looks fun :O

- jennychan12

oy.....!
-_-

- anonymous

|dw:1355977140299:dw|

- anonymous

|dw:1355977229842:dw|

- anonymous

|dw:1355977363994:dw|

- anonymous

the last sub. should be u=2cosht

- anonymous

@satellite73 @Zarkon

- anonymous

http://www.wolframalpha.com/input/?i=%28sinht%29^%285%2F3%29dt

- anonymous

I do not have enough information to solve this question..

- jennychan12

g is the antiderivative function if g(5) = 7, then g(1) = ?

- anonymous

why are giving this info now, you should mention this at the beginning.. come on..

- anonymous

|dw:1355996811040:dw|

- anonymous

http://www.freemathhelp.com/forum/archive/index.php/t-49380.html?s=81349e7c156e89af0e48100743b7ef6d

- anonymous

http://www.wolframalpha.com/input/?i=integrate+%28x^2%2B4x%29^%281%2F3%29dx+from+x%3D1+to+5

- jennychan12

thanks for trying to help. :)

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