Help please with integration by parts

**runner07** · October 28th 2007, 10:02 AM

it says

Calculate the given integral.
integral of [x^(7/2) * (ln(3x))] dx

i started it like this...
u= ln(3x) du= 1/x dx
v'= x^(7/2) v= (2/9)x^(9/2)

=ln(3x)(2/9 x^(9/2)) - 2/9[integral of x^(9/2) * (1/x) dx]

but now i'm stuck. please help

**Krizalid** · October 28th 2007, 10:30 AM

Take it more generally:

$\int x^a\ln(bx)\,dx,\,a>-1,\,b>0.$

The same procedure, integration by parts. Set $u=\ln(bx)$ & $dv=x^a\,dx.$

**runner07** · October 28th 2007, 11:19 AM

sorry, im not sure what you mean

**Jhevon** · October 28th 2007, 11:33 AM

Originally Posted by runner07

sorry, im not sure what you mean

you actually did what he said you should do. i don't see why you're stuck

$x^{9/2} \cdot \frac 1x = x^{9/2} \cdot x^{-1} = x^{9/2 - 1} = x^{7/2}$

and you can integrate that using the basic power rule

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