# Answer to Question #45962 in Integral Calculus for victor

Question #45962

Integrate with respect to x :

∫41x+1x√dx

∫41x+1x√dx

Expert's answer

∫(41x+x√x)dx=∫41xdx+∫x√xdx=41∫xdx+∫x

^{3/2}dx=41x^{2}/2+2x^{5/2}/5+C, where c is an arbitrary real constant.Need a fast expert's response?

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