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Integral calculus 2

posted by on .

This is also one I can't figure out:

∫ x^3 (2x^2 + 1)^1/2 dx

u is obviously = 2x^2 + 1
du = 4x dx
x dx = 1/4 du

1/4 ∫ x^3 u^1/2

from here I know I can't integrate because I can't have more than 1 variable.

Thank you

  • Integral calculus 2 - ,

    I rewrote it as

    ∫ ( x(2x^2+1)^(1/2) (x^2) dx

    I let u = x^2 and dv = x(2x^2 + 1)^(1/2) dx
    du/dx = 2x
    du = 2x dx

    v = (1/6)(2x^2 + 1)^(3/2)

    so we need: uv - ∫ v du

    uv
    = (1/6)(2x^2+1)^(3/2 (x^2)
    = (x^2/6)(2x^2 + 1)^(3/2)

    ∫ (1/6)(2x^2 + 1)^(3/2) (2x) dx
    = (1/30)(2x^2 + 1)^(5/2)

    finally our integral
    = (x^2/6)(2x^2 + 1)^(3/2) - (1/30)(2x^2 + 1)^(5/2)
    = (1/30)(2x^2 + 1)^(3/2) [ 5x^2 - (2x^2 + 1) ]
    = (1/30)(2x^2 + 1)^(3/2) (3x^2 - 1)

    wow, you better check my steps on this one.

  • Integral calculus 2 - ,

    thank you

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