What is #int x / (x+3)^(1/2) dx#?

1 Answer
George C.
Mar 11, 2018

#int x/(x+3)^(1/2) dx = 2/3(x+3)^(1/2)(x-6)+C#

Explanation:

#int x/(x+3)^(1/2) dx = int ((x+3)-3)/(x+3)^(1/2) dx#

#color(white)(int x/(x+3)^(1/2) dx) = int (x+3)^(1/2)-3(x+3)^(-1/2) dx#

#color(white)(int x/(x+3)^(1/2) dx) = 2/3(x+3)^(3/2)-6(x+3)^(1/2)+C#

#color(white)(int x/(x+3)^(1/2) dx) = 2/3(x+3)^(1/2)((x+3)-9)+C#

#color(white)(int x/(x+3)^(1/2) dx) = 2/3(x+3)^(1/2)(x-6)+C#

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