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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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