We can rewrite the integral as two fractions and then integrate teh em separately.
int((x-2)/(x^2+1))dx=color(red)(intx/(x^2+1)dx)-color(blue)(int2/(x^2+1)dx∫(x−2x2+1)dx=∫xx2+1dx−∫2x2+1dx
First integral
color(red)(intx/(x^2-2)dx)∫xx2−2dx
using int(f'(x))/f(x)dx=ln|f(x)|
d/(dx)(x^2+1)=2x
:.color(red)(intx/(x^2-2)dx=1/2ln|x^2+1|)
since x^2+1>0AA in RR " we can omit the modulus sign"
second integral
color(blue)(int2/(x^2+1)dx
we integrate by substitution
x=tanu=>dx=sec^2udu
color(blue)(int2/(x^2+1)dx=2int1/(tan^2u+1)sec^2udu
=color(blue)(2(int1/cancel(sec^2u)cancel(sec^2u)du)=2intdu
color(blue)(=2u=2tan^(-1)x)
putting the two results together we have:
int((x-2)/(x^2+1))dx=color(red)(intx/(x^2+1)dx)-color(blue)(int2/(x^2+1)dx
=color(red)(1/2ln(x^2+1))-color(blue)(2tan^(-1)x)+C
