integral of x arctan x dx

alba_ei

1. The problem statement, all variables and given/known data
[tex] \int x \arctan x \, dx [/tex]

3. The attempt at a solution
By parts,
[tex] u = \arctan x[/tex]
[tex] dv = x dx[/tex]
[tex] du = \frac{dx}{x^2+1}[/tex]
[tex]v = \frac{x^2}{2} [/tex]

[tex] \int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \frac{1}{2} \int \frac{x^2}{x^2+1} \, dx [/tex]

Again...by parts

[tex] u = x^2 [/tex]
[tex] dv = \frac{dx}{x^2+1} [/tex]
[tex] du = 2x dx [/tex]
[tex] v = arc tan x [/tex]

[tex]\int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \frac{x^2}{2}\arctan x - \int x \arctan x \, dx [/tex]
I back to the beginning, what did wrogn?

[tex]\int x \arctan x \, dx = - \int x \arctan x \, dx [/tex]

z-component

[tex]\int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \frac{x^2}{2}\arctan x - \int x \arctan x \, dx [/tex]

Add [tex]\int x \arctan x \, dx[/tex] to both sides, then solve for the integral, assuming your work is correct.

alba_ei

you mean like this? is the same, i back to the beginign

[tex]\int x \arctan x \, dx +\int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \frac{x^2}{2}\arctan x - \int x \arctan x \, dx +\int x \arctan x \, dx[/tex]

[tex]2\int x \arctan x \, dx = 0[/tex]

neutrino

Why use 'by parts' again? It would easier if you just add and subtract 1 from the numerator

scottie_000

why not try the substitution u=x^2+1 in that second integral...

Qyzren

for the integral x²/(x²+1)
you can rewrite it as (x² + 1 - 1)/(x²+1) => 1 - 1/(x²+1)

Gib Z

umm hmm, that leaves a nice (x - arctan x) for you there.

Neo139

http://www.maths.abdn.ac.uk/~igc/tch...nt/node34.html
Example 3.15