How to Solve the Integral of x arctan x dx?

[alba_ei]

Homework Statement

\int x \arctan x \, dx

The Attempt at a Solution

By parts,
u = \arctan x
dv = x dx
du = \frac{dx}{x^2+1}
v = \frac{x^2}{2}

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

Again...by parts

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

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

\int x \arctan x \, dx = - \int x \arctan x \, dx

 

[z-component]

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

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

 

[alba_ei]

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

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

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

\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

2\int x \arctan x \, dx = 0

 

[neutrino]

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

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/ma1002/int/node34.html
Example 3.15