How do you integrate $ln(x) x^ (3/2) dx$ ?

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Answer 1 · 2016-05-06T01:00:27

$intln(x)x^(3/2)dx=2/5x^(5/2)(ln(x)-2/5)+C$

For this problem, we will use integration by parts:

Let $u = ln(x)$ and $dv = x^(3/2)dx$

Then $du = 1/xdx$ and $v = 2/5x^(5/2)$

Applying the formula $intudv = uv - intvdu$ gives us

$intln(x)x^(3/2)dx = 2/5x^(5/2)ln(x)-int2/5x^(5/2)*1/xdx$

$=2/5x^(5/2)ln(x)-2/5intx^(3/2)dx$

$=2/5x^(5/2)ln(x) - 2/5(2/5x^(5/2))+C$

$=2/5x^(5/2)(ln(x)-2/5)+C$

How do you integrate ln(x) x^ (3/2) dxln(x) x^ (3/2) dx?