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Inverse Trig Integral

  1. Feb 23, 2008 #1
    I don't if its because I'm tired or what but I can't seem to integral the follow

    [tex]\int (-x+1)/(x^2+1)[/tex] I tried substitution, u=x^2+1 du=2x, doesn't appear to be anything there, u=-x+1, du=1, again doesn't appear to be anything there. The book shows it simply as tan^-1 (x), don't think its a partial fraction. Any ideas?
     
  2. jcsd
  3. Feb 23, 2008 #2
    [tex]\int\frac{1-x}{x^2+1}dx[/tex]

    Yes?

    [tex]\int\left(\frac{1}{x^2+1}-\frac{x}{x^2+1}\right)dx[/tex]

    How about now?
     
  4. Feb 23, 2008 #3
    blah, I should of seen that, thanks. :)
     
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