$\displaystyle

\frac{x - 3}{x^2 + 1}

$

This looks like a tan^-1x type answer but the actual answer is different and I am not sure of the process to get there.

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- Mar 7th 2009, 12:29 PMTYTYIntegral: have madness, need method
$\displaystyle

\frac{x - 3}{x^2 + 1}

$

This looks like a tan^-1x type answer but the actual answer is different and I am not sure of the process to get there. - Mar 7th 2009, 12:49 PMMentia
okay so separate it out:

$\displaystyle \frac{ x-3}{x^{2}+1 } = \frac{x }{ x^2+1} - \frac{ 3}{x^2+1 }$

then

$\displaystyle

\int_{ }^{ } \frac{ x-3}{x^{2}+1 }dx = \int_{ }^{ } \frac{x }{ x^2+1}dx - \int_{ }^{ } \frac{ 3}{x^2+1 }dx$

The part with the 3 is just arctan. Use a substitution on the other part:

u = x^2 + 1 --> du = 2xdx ---> dx = du/(2x), and so the integral should give a natural log. - Mar 7th 2009, 12:52 PMTYTY