Solving Inverse Trig Integral: Strategies and Tips

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The integral in question is ∫(-x+1)/(x^2+1)dx, which initially posed challenges for substitution methods. A successful approach involves breaking the integral into two parts: ∫(1/(x^2+1) - x/(x^2+1))dx. This simplification leads to the solution involving the arctangent function, specifically yielding tan^-1(x). The discussion highlights the importance of recognizing patterns in integrals and suggests that sometimes a fresh perspective can clarify the solution. Overall, the thread emphasizes effective strategies for solving inverse trigonometric integrals.
kuahji
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I don't if its because I'm tired or what but I can't seem to integral the follow

\int (-x+1)/(x^2+1) 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?
 
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\int\frac{1-x}{x^2+1}dx

Yes?

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

How about now?
 
blah, I should of seen that, thanks. :)
 

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