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How do you find the antiderivative of #x^(3/2)#?

1 Answer

Oct 20, 2016

#2/5 sqrt x^5# + c

Explanation:

#int x^(3/2) dx#

= #x^(3/2 + 1)/((3/2 +1)#

= #x^(5/2)/(5/2)#

= #2/5x^(5/2) + c#

= #2/5 sqrt x^5 # + c

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