# Thread: integration by parts help

1. ## integration by parts help

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

let u = tan^-1(x)
du = $\frac{1}{1+x^2}dx$

let v = $\frac{x^2}{2}$
dv = $x$

then got stuck after this
$\frac{x^2}{2}\tan^{-1}{x} - \int{\frac{x^2}{2}(\frac{1}{1+x^2})}dx$

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

let u = tan^-1(x)
du = $\frac{1}{1+x^2}dx$

let v = $\frac{x^2}{2}$
dv = $x$

then got stuck after this
$\frac{x^2}{2}\tan^{-1}{x} - \int{\frac{x^2}{2}(\frac{1}{1+x^2})}dx$
Doing great so far. Here's how to deal with the new integral.

forget the constant 1/2 you can pull out.

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

$= \int \left( 1 - \frac {1}{1 + x^2} \right)~dx$

Now continue

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

let u = tan^-1(x)
du = $\frac{1}{1+x^2}dx$

let v = $\frac{x^2}{2}$
dv = $x$

then got stuck after this
$\frac{x^2}{2}\tan^{-1}{x} - \int{\frac{x^2}{2}(\frac{1}{1+x^2})}dx$
Try this:
$\frac{x^2}{x^2 + 1} = \frac{x^2 + 1 - 1}{x^2 + 1} = 1 - \frac{1}{x^2 + 1}$

-Dan