$\displaystyle \int tan^{-1}u du $
is equal to
$\displaystyle utan^{-1}u-\frac{1}{2} ln(1+u^{2}) $
can somebody explain to me how this is true?
Here's your hint, as Plato suggested:
$\displaystyle \frac{d}{du} \left ( u~tan^{-1}(u)-\frac{1}{2} ln(1+u^{2}) \right )$
$\displaystyle = tan^{-1}(u) + \frac{u}{u^2 + 1} - \frac{1}{2} \left ( \frac{1}{1 + u^2} \cdot 2u \right )$
What does this tell you?
-Dan
the responses given seem to me to be geared towards verifying that the claim is true, rather than saying why it's true, or how it works. i believe what the poster was after is actually integrating arctan to get the result shown. in that case, we use integration by parts with $\displaystyle u = \arctan x$ and $\displaystyle dv = 1$ (you do know what i am referring to when i say $\displaystyle u \mbox{ and } dv$, right akhayoon?)