1. ## Integration by parts, help asap please, exam soon

$\int x tan^- (x) dx$ ( The Tan is inverse, i'm not sure how to properly display that in latex )

i did

$tan^- (x) (\frac{x^2}{2} ) - \int (\frac{x^2}{2}) (\frac{1}{1+x}) dx$

Is it right ? If yes

now how can I solve $\int (\frac{x^2}{2}) (\frac{1}{1+x}) dx$

2. The derivative of $\tan^{-1}{x}$ is $\frac{1}{1+x^2}$ and not $\frac{1}{1+x}$.

To solve $\frac{1}{2}\int \frac{x^2}{1+x^2}~dx$, simply add 1/subtract 1 (which is actually a shortcut around a polynomial division) to get:

$\frac{1}{2}\int \frac{\rlap{-------}1+x^2}{\rlap{-------}1+x^2}~dx - \frac{1}{2}\int \frac{1}{1+x^2}~dx$

3. Sorry if I sound a bit ignorant, but I don't really get it, I'm a bit new to all those stuff... could you explain a bit more? What did you add and what did you substract, and then what would i get ?

4. $\int (\frac{x^2}{2}) (\frac{1}{1+x}) dx$

I divided the polynomials in this integral and i got $\int 1 - \frac {1}{1+x^2}$

solution for that is x - ln | 1 + x^2 | + c

is this right ?

5. the integral of $\int \frac{1}{1+x^2}~dx
$
is $\tan^{-1}{x}
$

6. so $\int 1 - \frac {1}{1+x^2}$

=

x -

thanks a lot guys, eventhough i didn't understand Chop Suey method, but I'm glad i made some progress -,- just started learning integrals yesterday

Cheers