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The indefinate integral (or anti derivitave) of

x^4 Sqrt[5 + x^5] dx

Thanks alot for any help

2007-08-05 20:40:43 · 5 answers · asked by hurcut 1 in Science & Mathematics Mathematics

5 answers

notice d/dz of sqrt (5+x^5) is .5*(5+x^5)^-.5 * 5x^4

so basically x^4 is a chain rule derivative of the inside of your other expression. Let u = 5+x^5
du = 5*x^4 dx

so x^4(dx)= 1/5 * du

so you have 1/5 sqrt u du
which you can go from there

2007-08-05 20:47:25 · answer #1 · answered by Mutou Kazuki 2 · 0 0

I = ∫ x^4 / (5 + x^5)^(1/2) dx
Let u = 5 + x^5
du = 5x^4 dx
du / 5 = x^4 dx
I = (1 / 5) ∫ u^(1/2) du
I = (2 / 15) u^(3/2) + C
I = (2 / 15) (5 + x^5)^(3/2) + C

2007-08-05 22:13:37 · answer #2 · answered by Como 7 · 0 0

use chain rule...

else the fastest way is probably to ask ur tutor

2007-08-05 20:48:20 · answer #3 · answered by Anonymous · 0 0

http://mathworld.wolfram.com/

2007-08-05 20:43:39 · answer #4 · answered by Anonymous · 0 0

CHRIST MOUNTAIN FOR SQUIRTING (GIVE HIM FIVE WITH FIVE CHRIST MOUNTAIN) TO DX "youknowhim?"

2007-08-05 21:48:48 · answer #5 · answered by john_frank_evangelista 2 · 0 2

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