Hi,

Can someone tell me if this correct, please?

int. tan^-1xdx=xtan^-1x-1/2ln(1+x^2)+c

I used the formula "tan^-1udu=utan^-1u-1/2ln(1+u^2)+c

Thank you

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- Sep 7th 2007, 12:29 PMchocolateloverintegral using a table
Hi,

Can someone tell me if this correct, please?

int. tan^-1xdx=xtan^-1x-1/2ln(1+x^2)+c

I used the formula "tan^-1udu=utan^-1u-1/2ln(1+u^2)+c

Thank you - Sep 7th 2007, 01:56 PMKrizalid
- Sep 7th 2007, 01:57 PMJhevon
- Sep 7th 2007, 05:00 PMchocolateloverQuote:

Is it much to ask that you may use some of LaTeX or patentheses, pleaseż?

- Sep 7th 2007, 06:24 PMtopsquark
- Sep 7th 2007, 07:21 PMKrizalid
So, you can't, let's try!

By the mythic integration by parts formula, we have that

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

We gotta integrate $\displaystyle \arctan x$

So let's take $\displaystyle u=\arctan x\implies du=\frac1{1+x^2}\,dx$ & $\displaystyle dv=dx\implies v=x$, the integral becomes to

$\displaystyle \int\arctan x\,dx=x\arctan x-\int\frac x{1+x^2}\,dx$

Now the one million dollar question: could you take it from there? :D:D

__________

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