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Integral ∫x^3*√(x^25) dx 
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#1
Nov2512, 12:17 PM

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1. The problem statement, all variables and given/known data
∫x^3*√(x^25) dx 2. Relevant equations ∫u.dv=u.v∫du.v 3. The attempt at a solution So i tried to change the integral to ∫x*x^2*√(x^25)dx and u = x^25, then du = 2x, so 1/2*∫x^2*√(x^25) . Let u = √(x^25) , du = x/√(x^25) and dv = x^2 , v = x^3/3. Am I going in the right direction? Thanks in advance 


#2
Nov2512, 12:36 PM

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#3
Nov2512, 12:49 PM

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1/2*∫(u+5)*√udu so I multiplied and I got : 1/2*∫u^(3/2)+5u^(1/2) 1/2*1/5 * ∫u^(3/2)+u^(1/2) and I got u^(5/2)/25+(2u^(3/2))/3 and It's wrong :/ where did I failed? Thanks 


#4
Nov2512, 12:56 PM

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Integral ∫x^3*√(x^25) dx



#5
Nov2512, 12:59 PM

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The solution is supposed to be: x^3/3*(u^(3/2))2/15*(u^(5/2))+c 


#6
Nov2512, 01:08 PM

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#7
Nov2512, 01:11 PM

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http://www.wolframalpha.com/input/?i...t%28x%5E25%29 


#8
Nov2512, 01:21 PM

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#9
Nov2512, 01:24 PM

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#10
Nov2512, 01:31 PM

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