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Integration by Substitution

AP Calculus AB· difficulty 3/5

∫xx2+9 dx=\displaystyle\int x \sqrt{x^2 + 9}\,dx =

  • A

    2(x2+9)3/23+C\dfrac{2(x^2+9)^{3/2}}{3} + C

  • B

    (x2+9)3/22+C\dfrac{(x^2+9)^{3/2}}{2} + C

  • C

    (x2+9)3/2+C(x^2+9)^{3/2} + C

  • D

    (x2+9)3/23+C\dfrac{(x^2+9)^{3/2}}{3} + C

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Explanation

u=x2+9u = x^2 + 9, du=2x dxdu = 2x\,dx. ∫u du/2=u3/2/3\int \sqrt u\,du/2 = u^{3/2}/3.

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