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Integral of x arctan x dx

  • Thread starter alba_ei
  • Start date
  • #1
38
1

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]
 
Last edited:

Answers and Replies

  • #2
478
2
[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.
 
  • #3
38
1
[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]
 
  • #4
2,063
2
[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
 
  • #5
why not try the substitution u=x^2+1 in that second integral...
 
  • #6
44
0
for the integral x²/(x²+1)
you can rewrite it as (x² + 1 - 1)/(x²+1) => 1 - 1/(x²+1)
 
  • #7
Gib Z
Homework Helper
3,346
5
umm hmm, that leaves a nice (x - arctan x) for you there.
 

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