Quote:
Originally posted by Thisisit
∫ (x^2 + 12x + 27)^-0.5 . dx
(that's everything in the brackets to the power of -0.5, in case the notation is unclear)
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Quote:
Originally posted by TheRoofInhabitant
No one's tried omg
It's too long to explain here though, I made it with the substitution method, with t - x = (x² + 12x + 27)⁰ˑ⁵ and in the end the result is ln(2(x² + 12x + 27)⁰ˑ⁵ + 2x + 27) + c
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that's an interesting way of going about it, I think substitution with hyperbolic functions is an easier way though
∫ (x^2 + 12x + 27)^-0.5 . dx
= ∫ [(x+6)^2 -9]^-0.5 . dx
x+6 = 3cosh(u)
dx/du = 3sinh(u)
∫ [ (x+6)^2 -9]^-0.5 . dx
= ∫ 3sinh(u)*(9cosh^2(u) -9)^-0.5 . du
= ∫ 3sinh(u)*(9sinh^2(u))^-0.5 . du
= ∫ 3sinh(u)*(3sinh(u))^-1 . du
= ∫ 1 .du
= u + c
rearrange x+6 = 3cosh(u) to get u in terms of x
u = arcosh((x+6)/3)
thus
∫ (x^2 + 12x + 27)^-0.5 . dx = arcosh((x+6)/3) + c