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

1 Answer

Oct 14, 2015

#2/3sqrt(x^3)+3lnx-4e^x+C#

Explanation:

Rewriting in a different form using laws of exponents and surds, and then integrating term by term using normal rules of integration, we get :

#int(x^(1/2)+3*1/x-4e^x)dx#

#=(x^(3/2))/(3/2)+3lnx-4e^x+C#

#=2/3sqrt(x^3)+3lnx-4e^x+C#

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