How do you integrate #int (4x)/sqrt(3x)# using substitution?

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Answer 1 · 2016-08-01T18:15:15

I found: #8/27(3x)^(3/2)+c#

Have a look:
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Answer 2 · 2016-08-01T18:42:11

#(8x^(3/2))/3^(3/2)+C#

However, we should see that substitution is a waste of time:

#int(4x)/sqrt(3x)dx=4/sqrt3intx/sqrtxdx=4/sqrt3intx^(1/2)dx#

Using the rule #intx^ndx=x^(n+1)/(n+1)+C#. where #n!=-1#, this becomes:

#=4/sqrt3(x^(3/2)/(3/2))+C=8/(3sqrt3)x^(3/2)+C=(8x^(3/2))/3^(3/2)+C#