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How do you find the indefinite integral of #int (x^4-6x^3+e^xsqrtx)/sqrtx dx#?

1 Answer

Cesareo R. · mason m

Sep 27, 2016

#2/9x^(9/2)-12/7x^(7/2)+e^x+C#

Explanation:

#int (x^4-6x^3+e^xsqrtx)/sqrtx dx = int (x^(7/2)-6x^(5/2)+e^x)dx#
#=1/(7/2+1)x^(7/2+1)-6/(5/2+1)x^(5/2+1)+e^x+C#

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