Posted by laura on Saturday, December 29, 2012 at 8:16am.
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)
oops. secθ = √(4+x^2)/2
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