What is the antiderivative of #x sqrtx#?

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Answer 1 · 2017-01-17T20:09:35

#2/5x^2sqrtx+C#

Note that #xsqrtx=x^1x^(1/2)=x^(1+1/2)=x^(3/2)#. Then:

#intxsqrtxcolor(white).dx=intx^(3/2)dx#

Now use the rule #intx^ndx=x^(n+1)/(n+1)+C# we see that:

#=x^(3/2+1)/(3/2+1)+C=x^(5/2)/(5/2)+C=2/5x^(5/2)+C#

If you so desire, you can write that

#x^(5/2)=x^(4/2)x^(1/2)=x^2sqrtx#

So:

#intxsqrtxcolor(white).dx=2/5x^2sqrtx+C#