Integrate (sqr(x-4))/x?

Hello,

∫ {[√(x - 4)] /x} dx =

let:

√(x - 4) = u

x - 4 = u²

x = u² + 4

dx = 2u du

yielding, by substitution:

∫ {[√(x - 4)] /x} dx = ∫ [u /(u² + 4)] 2u du =

∫ 2 [u² /(u² + 4)] du =

being numerator and denominator of the same order, let's add and subtract 4 in the numerator:

∫ 2 {[(u² + 4) - 4] /(u² + 4)} du =

(distributing and simplifying)

∫ 2 {[(u² + 4) /(u² + 4)] - [4 /(u² + 4)]} du =

∫ 2 {1 - [4 /(u² + 4)]} du =

∫ {2 - [8 /(u² + 4)]} du =

(splitting into two integrals and factoring constants out)

2 ∫ du - 8 ∫ [1 /(u² + 4)] du =

2u - 8 ∫ [1 /(u² + 4)] du =

let's write the denominator of the remaining integrand as a sum of squares:

2u - 8 ∫ [1 /(u² + 2²)] du =

let's divide numerator and denominator by 2² (to turn the denominator into the form

f(u)² + 1):

2u - 8 ∫ {(1/2²) /[(u²/2²) + (2²/2²)]} du =

2u - 8 ∫ {(1/2)² /[(u/2)² + 1]} du =

let's factor out 1/2, obtaining in the numerator the derivative of u/2 (that is just 1/2):

2u - 8(1/2) ∫ (1/2) du /[(u/2)² + 1] =

2u - 4 ∫ d(u/2) /[(u/2)² + 1] =

(being this integral of the form ∫ d[f(x)] /[f(x)² + 1] = arctan[f(x)] + C)

2u - 4arctan(u/2) + C

finally let's substitute back for u:

2√(x - 4) - 4arctan{[√(x - 4)] /2} + C =

ending with:

2√(x - 4) - 4arctan{[√[(x - 4)/4]} + C (the denominator 4 in the argument of the arctangent is inside the root)

I hope it's helpful