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Integration by Substitution

AP Calculus AB· difficulty 3/5

∫012x(x2+1)3 dx=\displaystyle\int_0^1 2x(x^2+1)^3\,dx =

  • A

    44

  • B

    14\tfrac{1}{4}

  • C

    154\tfrac{15}{4}

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  • D

    164\tfrac{16}{4}

Explanation

u=x2+1u = x^2 + 1. When x=0,u=1x = 0, u = 1; x=1,u=2x = 1, u = 2. ∫12u3 du=[u4/4]12=4−1/4=15/4\int_1^2 u^3\,du = [u^4/4]_1^2 = 4 - 1/4 = 15/4.

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