integration on inverse trig

**akhayoon** · January 31st 2008, 03:21 PM

$\int tan^{-1}u du$

is equal to

$utan^{-1}u-\frac{1}{2} ln(1+u^{2})$

can somebody explain to me how this is true?

**Plato** · January 31st 2008, 03:50 PM

Originally Posted by akhayoon

$\int tan^{-1}u du$ is equal to
$utan^{-1}u-\frac{1}{2} ln(1+u^{2})$ can somebody explain to me how this is true?

Did you differentiate the answer?
Do it!
That shows you how it works.

**akhayoon** · January 31st 2008, 03:55 PM

well thats the thing I haven't been taught these

it's just given to me as a formula...but I would like to understand how it happened so that way I dont have to memorize it...

thanks

**Plato** · January 31st 2008, 04:02 PM

Originally Posted by akhayoon

well thats the thing I haven't been taught these
it's just given to me as a formula...but I would like to understand how it happened so that way I dont have to memorize it...

Sorry to say, but I find that notion ridiculous.

**akhayoon** · January 31st 2008, 04:07 PM

well..I really don't care you how you find that notion...all I really want is a hint on how to begin solving that problem really, which is what I always get when I have a question on this forum.

**topsquark** · January 31st 2008, 04:46 PM

Originally Posted by akhayoon

$\int tan^{-1}u du$

is equal to

$utan^{-1}u-\frac{1}{2} ln(1+u^{2})$

can somebody explain to me how this is true?

Originally Posted by Plato

Did you differentiate the answer?
Do it!
That shows you how it works.

Here's your hint, as Plato suggested:
$\frac{d}{du} \left ( u~tan^{-1}(u)-\frac{1}{2} ln(1+u^{2}) \right )$

$= tan^{-1}(u) + \frac{u}{u^2 + 1} - \frac{1}{2} \left ( \frac{1}{1 + u^2} \cdot 2u \right )$

What does this tell you?

-Dan

**Jhevon** · January 31st 2008, 07:20 PM

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 $u = \arctan x$ and $dv = 1$ (you do know what i am referring to when i say $u \mbox{ and } dv$ , right akhayoon?)

**Krizalid** · January 31st 2008, 07:24 PM

A similar integration by parts reasoning is useful here: take the following function $f(u)=u\arctan u.$ Contemplate its derivative and integrate, the rest follows.

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