How do you find the integral $ln (x) x^{\frac{3}{2}} d x$ $ln(x) x^(3/2) dx$ ?

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How do you find the integral $ln (x) x^{\frac{3}{2}} d x$ $ln(x) x^(3/2) dx$ ?

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Answer 1 · 2015-06-18T17:35:23

To do integration by parts:

$u v = \int v d u$ $uv = intvdu$

I would pick a $u$ $u$ that gives me an easy time differentiating, and a $d v$ $dv$ that I would easily be able to integrate multiple times if necessary. If the same integral comes back, or variables aren't going away, either it's cyclic or you should switch your $u$ $u$ and $d v$ $dv$ .

Let:
$u = ln x$ $u = lnx$
$d u = \frac{1}{x} d x$ $du = 1/xdx$
$d v = x^{\frac{3}{2}} d x$ $dv = x^(3/2)dx$
$v = \frac{2}{5} x^{\frac{5}{2}}$ $v = 2/5x^(5/2)$

$\Rightarrow \frac{2}{5} x^{\frac{5}{2}} ln x - \int \frac{2}{5} (\frac{x^{\frac{5}{2}}}{x}) d x$ $=> 2/5x^(5/2)lnx - int2/5(x^(5/2)/x)dx$

Nice! That's not too bad.

$= \frac{2}{5} x^{\frac{5}{2}} ln x - \int \frac{2}{5} x^{\frac{3}{2}} d x$ $= 2/5x^(5/2)lnx - int2/5x^(3/2)dx$

$= \frac{2}{5} x^{\frac{5}{2}} ln x - \frac{4}{25} x^{\frac{5}{2}} + C$ $= 2/5x^(5/2)lnx - 4/25x^(5/2) + C$

$= \frac{10}{25} x^{\frac{5}{2}} ln x - \frac{4}{25} x^{\frac{5}{2}} + C$ $= 10/25x^(5/2)lnx - 4/25x^(5/2) + C$

$= x^{\frac{5}{2}} [\frac{10}{25} ln x - \frac{4}{25}] + C$ $= x^(5/2)[10/25lnx - 4/25] + C$

$= \frac{2}{25} x^{\frac{5}{2}} [5 ln x - 2] + C$ $= 2/25x^(5/2)[5lnx - 2] + C$

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