Integral question using substitution?

Question

The integral of ((x-3)/x^5)^1/2 dx

Answer 1

Maybe I'm blind, but i can't see a simple substitution.

But there is a standard procedure to solve such integrals.

If you can transform the integrand to a rational function of x and the n-th root of a rational function linear in x like [(ax + b)/(cx + d)]^(1/n), you get a rational integrand by substituting t = [(ax + b)/(cx + d)]^(1/n).

∫ ((x-3)/x⁵)^(1/2) dx

factor out x⁴ from denominator to get a linear term

= ∫ (1/x⁴)^(1/2) ∙ ((x-3)/x)^(1/2) dx

= ∫ (1/x²) ∙ ((x-3)/x)^(1/2) dx

next substitute

t = ((x-3)/x)^(1/2)

<=>

t² = (x-3)/x

<=>

x = 3/(1-t²)

=>

dx = 6t/(1-t²)²

= ∫ (1/ (3/(1-t²))²) ∙ t ∙ (6t/(1-t²)) dt

= ∫ ((1-t²)²/9) ∙ t ∙ (6t/(1-t²)² dt

= ∫ (2/3) ∙ t²  dt

= (2/9) ∙ t³ + c

back substitution t = ((x-3)/x)^(1/2)

= (2/9) ∙ ((x-3)/x)^(3/2) + c