# Calculus

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∫[(x+3)/(x^2 +9)]=????? Can you give me a hint for solving it?

• Calculus -

I 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)

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