Inverse Trig Integral

kuahji

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?

rocomath

[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?

kuahji

blah, I should of seen that, thanks. :)