How Does the Denominator x+a Become 1-t in Integral Substitution?

[Belgium 12]

Hi members,

see attached Pdf file.
For a<0,write b=-a and let x=bt.Then

My question:

how becomes the denominator x+a to 1-t? I don't see it

Thank you

 

[HallsofIvy]

The integral is
\int_0^\infty \frac{x^{\mu-1}dx}{x+ a}

You attachment says "if a< 0 let b= -a" so becomes
\int_0^\infty \frac{x^{\mu-1}dx}{x- b}

and "let x= bt". The denominator is x- b= bt- b= b(t- 1)

Factor out "-b" leave 1- t in the denominator,