Exercises on integrals

  1. ∫ (3x + 1)3 dx

  2. ∫ (3x2 + 1)3 dx

  3. ∫ x(3x2 + 1)3 dx

  4. ∫ (x5/6 + 2x

    −2 −3x−1 + 2) dx

Solutions

  1. Substitute s = 3x + 1; ds = 3 dx and solve 1/3 ⋅ ∫ s3 ds = 1/12 ⋅ s4 = 1/12 ⋅ (3x + 1)4 + const. ■

  2. We use the binomial theorem (3x2 + 1)3 = 27x6 + 27 x4 + 9x2 + 1. So

    ∫ (27x6 + 27 x4 + 9x2 + 1) dx = 27/7 x7 + 27 x5 + 3x3 + x+ constant.  ■

  3. Substitute s = 3x2 + 1; ds = dx/6 we have 1/6 ⋅ ∫ s3 dx = s4/24; Substituting back we have (3x2 + 1)4/24 = 27/8 ⋅ x8 + 9/2 ⋅ x6 + 9/4 ⋅ x4 + x2/2 + constant. ■

  4. 6/11 ⋅ x11/6 + 2x − 2/x − 3log x + constant.  ■

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