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integrate by parts

integrate (4+x^2)^1/2

  • maths -

    I'd do a trig substitution first:

    x = 2tanθ
    4+x^2 = 4+tan^2θ = 4sec^2θ
    dx = 2sec^2θ dθ

    and the integral is now

    ∫2secθ * 2sec^2θ dθ
    ∫4sec^3θ dθ

    now we can do the integration by parts. Let
    u = secθ, du = secθtanθ dθ
    dv = sec^2θ dθ, v = tanθ

    ∫u dv = uv - ∫v du
    = 4secθtanθ - 4∫secθtan^2θ dθ
    = 4secθtanθ - 4∫secθ(sec^2θ-1)dθ
    = 4secθtanθ - 4∫sec^3θ dθ + 4∫secθ dθ

    If we let I = ∫4sec^3θ dθ, we now have

    I = 4secθtanθx - I + 4ln(secθ+tanθ)
    2I = 4secθtanθ + 4ln(secθ+tanθ)
    I = 2secθtanθ + 2ln(secθ+tanθ)

    Now substitute back in for θ and we have

    θ = arctan(x/2)
    tanθ = x/2
    secθ = 2/√(4+x^2)

    ∫√(4+x^2) dx = 2*√(4+x^2)/2*x/2 + 2ln(x/2 + √(4+x^2))
    = x/2 √(4+x^2) + 2arcsinh(x/2)

  • maths -

    oops. secθ = √(4+x^2)/2

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