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- Mar 11th 2010, 01:27 PMstache31help with integral by parts
- Mar 11th 2010, 01:32 PMe^(i*pi)
$\displaystyle u' = \frac{1}{x}$

$\displaystyle v = \frac{46}{3}x^3$

$\displaystyle \int u\,dv = uv - \int v\,du$

Your integral is therefore equal to

$\displaystyle \frac{46}{3}x^3 \cdot \ln(x) - \int \left(\frac{46}{3}x^3 \cdot \frac{1}{x}\right)\,dx$

Use algebra to simplify the term inside the brackets, then evaluate the integral which is a standard power rule one - Mar 11th 2010, 01:38 PMstache31
thanks man. also, rammstein is awesome!