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Answer to Question #45962 in Integral Calculus for victor

Question #45962
Integrate with respect to x :
∫41x+1x√dx
1
Expert's answer
2014-09-11T13:41:47-0400
∫(41x+x√x)dx=∫41xdx+∫x√xdx=41∫xdx+∫x3/2dx=41x2/2+2x5/2/5+C, where c is an arbitrary real constant.

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