Integration by parts

**drew.walker** · September 7th 2009, 08:38 PM

Using integration by parts, would the following be correct?

$\int(2x-1)\ln x\,dx$
$u = \ln x \Rightarrow du = \frac{1}{x}\,dx$
$dv = (2x - 1)\,dx \Rightarrow v = \int 2x - 1\,dx = x^2 - x$

$\int(2x - 1)\ln x\,dx = \ln x(x^2 - x) - \int{\frac{x^2 - x}{x}}\,dx$

$= \ln x(x^2 - x) - \int{x - 1}\,dx$
$= \ln x(x^2 - x) - \frac{x^2}{2} + x + C$

**JG89** · September 7th 2009, 08:42 PM

It's fine. Remember, you could always differentiate the anti-derivative that you arrive at and see if it returns the integrand.

**drew.walker** · September 7th 2009, 08:49 PM

Ah, good point. Thanks.

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