How do I integrate (x-1)/(16+x^2)

note:split into two integrals, and use substitution on the first)

yeah, it seems like you DO know how to integrate this one
your 2 integrals are:

∫ x / 16 + x^2 dx - ∫ 1 / 16 + x^2 dx

for the first one, substitute u = 16 + x^2 and du = 2x dx (so you have to multiply the integral by 1/2)
= 1/2 ∫ du/u = 1/2 ln(u) = 1/2 ln(16 + x^2) + c

for the second integral, put 16 as (4)^2
and the answer will be = 1/4 tan^-1(x/4) + c

note:
∫ 1 / a^2 + x^2 dx = 1/a tan^-1(x/a) + c

hope it helped!

(x-1)/(x^2+16)

= x/(x^2+16) - 1/(x^2+16)

as d/dx (x^2+16) = 2x integral of 1st one is 1/2 ln (x^2+ 16) ( no need to keep mod as x^2+16 > 0

and integral of 1/(x^2 + 16) = 1/4 arctan (x/4)

so we get the result

split it as (x/x^2+16)-(1/x^2+16)
then perform integration. the first one is in arcsin form and the second one is in arc tan form.

manipulate the denominator until its differentiated form is x-1