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May 1, 2016
Posted by **Vivian** on Wednesday, February 8, 2012 at 1:35am.

- Calculus -
**Reiny**, Wednesday, February 8, 2012 at 8:23amI split (x+3)/(x^2+9) into

x/(x^2 + 9) + 3/(x^2+9)

= (1/2)(2x/(x^2+9) + 3/(x^2 + 9)

the integral of the first is (1/2) ln(x^2+9)

the second I looked up in my old integral tables and I would get

tan^2 (x/3)

so ∫[(x+3)/(x^2 +9)] dx

= (1/2)ln(x^2+9) + tan^1 (x/3)