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∫ dx / (x√(4x⁴ - 25))?
Here's what I tried. I asked this question because the answers in Wolfram Alpha and Symbolab both had some weird arctan answer, which I know can be adjusted from another inverse trig expression, but it's not the "natural" way someone would do it by hand since it doesn't fit the arctan derivative:
Let u = 2x²
du/dx = 4x → dx = du/4x
Partially in terms of u, the integral is
∫ du /[4x∙x√(u²-25)]
= ∫ du /[4x²√(u²-25)]
=½∫ du /[2x²√(u²-25)]
= ½∫ du /[ |u|√(u²-25)]
= 1/10 arcsec (2x²/5) + C
2 Answers
- germanoLv 76 years ago
Hello,
∫ dx /[x √(4x⁴ - 25)] =
first let's multiply and divide by x³:
∫ x³ dx /[x⁴ √(4x⁴ - 25)] =
let:
√(4x⁴ - 25) = u
4x⁴ - 25 = u²
4x⁴ = u² + 25
x⁴ = (u² + 25)/4
(differentiating both sides)
d(x⁴) = d[(u² + 25)/4]
4x³ dx = [(2u)/4] du
4x³ dx = (u/2) du
x³ dx = (1/4)(u/2) du = (u/8) du
yielding, by substitution:
∫ x³ dx /[x⁴ √(4x⁴ - 25)] = ∫ (u/8) du /{[(u² + 25)/4] u} =
∫ (u/8) {4 /[u (u² + 25)} du =
(simplifying and factoring the constant out)
(1/2) ∫ du /(u² + 25) =
let's write the denominator as a sum of squares:
(1/2) ∫ du /(u² + 5²) =
let's divide numerator and denominator by 5² to turn the denominator into the form f(u)² + 1:
(1/2) ∫ (1/5²) du /[(u²/5²) + (5²/5²)] =
(1/2) ∫ (1/5)² du /[(u/5)² + 1] =
let's factor out 1/5, obtaining the derivative of u/5 (that is 1/5) in the numerator:
(1/2)(1/5) ∫ (1/5) du /[(u/5)² + 1] =
(1/10) ∫ d(u/5) /[(u/5)² + 1] =
(being this of the form ∫ d[f(x)] /[f(x)² + 1] = arctan[f(x)] + C)
(1/10)arctan(u/5) + C
let's substitute back √(4x⁴ - 25) for u, ending with:
(1/10)arctan{[√(4x⁴ - 25)] /5} + C (I've just checked this successfully)
I hope it's helpful
