∫ dx / (x√(4x⁴ - 25))?

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

-(1/10) ArcTan[5/Sqrt[-25 + 4 x^4]]