# Thread: integral using a table

1. ## integral 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

2. Originally Posted by chocolatelover
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
Is it much to ask that you may use some of LaTeX or patentheses, pleaseż?

Actually, this it's solved applyin' integration by parts. Show your attempt.

3. Originally Posted by chocolatelover
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
it's correct. please use the appropriate parentheses

EDIT: I know you were required to use an integration table to find the answer, but Krizalid may be on to something. try doing it by parts--for practice

4. Is it much to ask that you may use some of LaTeX or patentheses, pleaseż?
Sorry. I've tried it, but I can't get it to work.

5. Originally Posted by chocolatelover
Sorry. I've tried it, but I can't get it to work.
You can't get parenthesis to work???

All we ask is some way to distinguish things like sq. 1 - x^2 from sq. (1 - x^2).
-Dan

6. Originally Posted by chocolatelover
Sorry. I've tried it, but I can't get it to work.
So, you can't, let's try!

By the mythic integration by parts formula, we have that

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

We gotta integrate $\arctan x$

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

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

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

__________

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