Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 40

intx/sqrt(x^2-6x+5)dx
Let's complete the square of the denominator of the integrand,
=intx/sqrt((x-3)^2-4)dx
Now apply integral substitution:u=x-3
=>du=1dx
=int(u+3)/sqrt(u^2-2^2)du
Now apply the sum rule,
=intu/sqrt(u^2-2^2)du+int3/sqrt(u^2-2^2)du
Now let's evaluate the first integral ,
intu/sqrt(u^2-4)du
Apply integral substitution:v=u^2-4
=>dv=2udu
=int1/sqrt(v)(dv)/2
Take the constant out and apply the power rule,
=1/2(v^(-1/2+1)/(-1/2+1))
=1/2(2/1)v^(1/2)
=sqrt(v)
Substitute back v=u^2-4
=sqrt(u^2-4)
Now let's evaluate the second integral,
int3/sqrt(u^2-2^2)du
Take the constant out,
=3int1/sqrt(u^2-2^2)du
Use the standard integral:int1/sqrt(x^2-a^2)dx=ln|x+sqrt(x^2-a^2)|
=3ln|u+sqrt(u^2-2^2)| ,
So add the result of the two integrals,
sqrt(u^2-4)+3ln|u+sqrt(u^2-4)|
Substitute back u=x-3 and add a constant C to the solution,
=sqrt((x-3)^2-4)+3ln|x-3+sqrt((x-3)^2-4)|+C
=sqrt(x^2-6x+9-4)+3ln|x-3+sqrt(x^2-6x+9-4)|+C
=sqrt(x^2-6x+5)+3ln|x-3+sqrt(x^2-6x+5)|+C