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

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- Oct 28th 2007, 10:02 AMrunner07Help please with integration by parts
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 - Oct 28th 2007, 10:30 AMKrizalid
Take it more generally:

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

The same procedure, integration by parts. Set $\displaystyle u=\ln(bx)$ & $\displaystyle dv=x^a\,dx.$ - Oct 28th 2007, 11:19 AMrunner07
sorry, im not sure what you mean

- Oct 28th 2007, 11:33 AMJhevon