How to Solve the Integral of x arctan x dx?

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Homework Statement


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

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]
 
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[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.
 
z-component said:
[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.

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]
 
alba_ei said:
[tex]- \frac{1}{2} \int \frac{x^2}{x^2+1} \, dx[/tex]
Why use 'by parts' again? It would easier if you just add and subtract 1 from the numerator
 
why not try the substitution u=x^2+1 in that second integral...
 
for the integral x²/(x²+1)
you can rewrite it as (x² + 1 - 1)/(x²+1) => 1 - 1/(x²+1)
 
http://www.maths.abdn.ac.uk/~igc/tch/ma1002/int/node34.html
Example 3.15
 
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