How do you integrate 1/((x^2)(sqrt(x^2 + 8))?

Let x = (√8) tanθ, then
dx = (√8) sec²θ and sinθ = x / √(x² + 8)

∫{1 / [x² √(x² + 8)]} dx
= ∫{1 / [8 tan²θ √(8 tan²θ + 8)]} (√8) sec²θ dθ
= (1/8)∫{1 / [tan²θ √(tan²θ + 1)]} sec²θ dθ
= (1/8)∫{1 / [tan²θ secθ]} sec²θ dθ
= (1/8)∫{secθ / tan²θ} dθ
= (1/8)∫{(1/cosθ) / (sin²θ/cos²θ} dθ
= (1/8)∫{cosθ / sin²θ} dθ
= (1/8)∫{1 / sin²θ} (cosθ dθ)
= (1/8)∫{1 / sin²θ} (d sinθ)
= (1/8) (- 1 / sinθ) + C
= (1/8) {-[√(x² + 8)] / x} + C
= -{[√(x² + 8)] / (8x)} + C

For sqrt(x^2+8) you must recall trigonometric identities,

∫ 1/((x^2)(sqrt(x^2 + 8))) dx

let x = sqrt(8)tan(u) , and dx = 2sqrt(2)sec^2(u) du , and sqrt(x^2+8) = sqrt(8(tan^2(u)+1)) , and recall that sec^(x) = 1 + tan^2(x) for all x € R, hence,

sqrt(x^2+8) = 2sqrt(2)sqrt(sec^2(u)) = 2sqrt(2)sec(u), and we get

= ∫ (1 / (2sqrt(2)tan^2(u)2sqrt(2)sec(u)))(2sqrt(2)sec^2(u)) du , simplify the integrand,

= ∫ (1 / 2sqrt(2)tan^2(u))(sec(u)) du

= (1/2sqrt(2) ∫ (cos(u)/sin^2(u)) du , we get

= (1/2sqrt(2)) ∫ cos(u)csc^2(u) du , and

= (1/2sqrt(2)) (-csc(u) + c) ,subtitute back to get the result in terms of x