# maths

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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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