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Integration by Parts: Int: x*arctan(x) dx

  • Thread starter MathHawk
  • Start date
  • #1
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Because of circumstance (my desire to graduate in 5 years or less), I've been forced to attempt Calc 2 in 2 months time online over the summer. About 75% of it is going smoothly (compared with 105% or so of Calc 1).

Homework Statement



I'm to solve the indefinite integral: [tex]\int[/tex] x * arctan(x) dx



Homework Equations


Integration by parts is done using: [tex]\int[/tex]u dv = uv - [tex]\int[/tex]v du


The Attempt at a Solution


It seems pretty obvious that u = arctan(x) and that dv = x. From this, du = [tex]\frac{1}{1+x^2}[/tex] dx and v = [tex]\frac{1}{2}[/tex]x2.

Using the integration by parts formula:

[tex]\int[/tex] x * arctan(x) dx = [tex]\frac{1}{2}[/tex]x2arctan(x) - [tex]\frac{1}{2}[/tex][tex]\int[/tex][tex]\frac{x^2}{1 + x^2}[/tex]dx



Now integration by parts must be used again. It seems obvious to select u = x2, dv = [tex]\frac{1}{1 + x^2}[/tex]. du = 2xdx, dv = arctan(x).



[tex]\int[/tex] x * arctan(x) dx = [tex]\frac{1}{2}[/tex]x2arctan(x) - [tex]\frac{1}{2}[/tex] ( x2arctan(x) - 2 [tex]\int[/tex]xarctan(x) ).

Simplified:

[tex]\int[/tex] x * arctan(x) dx = 0 + [tex]\int[/tex] x * arctan(x) dx.



While this is very true, it doesn't help me find the integral. Switching my u and dv in either use of the integration by parts formula hasn't yielded a solution for me in my attempts yet. Thank you in advance for your help :smile:.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
[tex]\frac{x^2}{x^2+1}=\frac{x^2+1 -1}{x^2+1}=1-\frac{1}{x^2+1}[/tex]
 
  • #3
9
0
You are my Bokonon, only you tell truths.
 
  • #4
9
0
Thank you, in other words.
 

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