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:

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

What is the antiderivative of x sqrtxx sqrtx?