Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 14

int(5x-2)/(x-2)^2dx
Let's use partial fraction decomposition on the integrand,
(5x-2)/(x-2)^2=A/(x-2)+B/(x-2)^2
5x-2=A(x-2)+B
5x-2=Ax-2A+B
comparing the coefficients of the like terms,
A=5
-2A+B=-2
Plug in the value of A in the above equation,
-2(5)+B=-2
-10+B=-2
B=-2+10
B=8
So now int(5x-2)/(x-2)^2dx=int(5/(x-2)+8/(x-2)^2)dx
Now apply the sum rule,
=int5/(x-2)dx+int8/(x-2)^2dx
Take the constant's out,
=5int1/(x-2)dx+8int1/(x-2)^2dx
Now let's evaluate each of the above two integrals separately,
int1/(x-2)dx
Apply integral substitution u=x-2
=>du=dx
=int1/udu
Use the common integral :int1/xdx=ln|x|
=ln|u|
Substitute back u=x-2
=ln|x-2|
Now evaluate the second integral,
int1/(x-2)^2dx
Apply integral substitution:v=x-2
dv=dx
=int1/v^2dv
=intv^(-2)dv
Apply the power rule,
=v^(-2+1)/(-2+1)
=-v^(-1)
=-1/v
Substitute back v=x-2
=-1/(x-2)
:.int(5x-2)/(x-2)^2dx=5ln|x-2|+8(-1/(x-2))
Add a constant C to the solution,
=5ln|x-2|-8/(x-2)+C