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July 30, 2014

July 30, 2014

Posted by **laura** on Saturday, December 29, 2012 at 8:16am.

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

- maths -
**Steve**, Saturday, December 29, 2012 at 12:53pmI'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 -
**Steve**, Saturday, December 29, 2012 at 12:55pmoops. secθ = √(4+x^2)/2

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