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Integral of x arctan x dx

by alba_ei
Tags: arctan, integral
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alba_ei
#1
Mar25-07, 05:11 PM
P: 38
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]
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z-component
#2
Mar25-07, 05:26 PM
z-component's Avatar
P: 483
[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
#3
Mar25-07, 05:43 PM
P: 38
Quote Quote by z-component View Post
[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]

neutrino
#4
Mar26-07, 01:29 AM
P: 2,046
Integral of x arctan x dx

Quote Quote by alba_ei View Post
[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
scottie_000
#5
Mar26-07, 04:24 AM
P: 49
why not try the substitution u=x^2+1 in that second integral...
Qyzren
#6
Mar26-07, 04:49 AM
P: 43
for the integral x/(x+1)
you can rewrite it as (x + 1 - 1)/(x+1) => 1 - 1/(x+1)
Gib Z
#7
Mar26-07, 07:07 AM
HW Helper
Gib Z's Avatar
P: 3,348
umm hmm, that leaves a nice (x - arctan x) for you there.
Neo139
#8
Dec1-08, 09:10 AM
P: 1
http://www.maths.abdn.ac.uk/~igc/tch...nt/node34.html
Example 3.15


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