What is #int (x+3) / sqrt(x)dx#?

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Answer 1 · 2015-12-14T07:26:15

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

Using basic properties of integration, along with that #intx^kdx = x^(k+1)/(k+1)+C# for #k != 0#, we have

#int(x+3)/sqrt(x)dx = int(x/sqrt(x) + 3/sqrt(x))dx#

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

#= intx^(1/2)dx + int3x^(-1/2)dx#

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

#= x^(3/2)/(3/2) + 3x^(1/2)/(1/2) + C#

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