Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 33

intx/(x^2-6x+10)^2dx
Let's rewrite the integrand as,
=1/2int(2x)/(x^2-6x+10)^2dx
=1/2int(2x-6+6)/(x^2-6x+10)^2dx
=1/2[int(2x-6)/(x^2-6x+10)^2dx+int6/(x^2-6x+10)^2dx] --------------------(1)
Now let' evaluate each of the above two integrals separately,
int(2x-6)/(x^2-6x+10)^2dx
Let's apply integral substitution:u=x^2-6x+10
=>du=(2x-6)dx
=int1/u^2du
=intu^(-2)du
Now from the integer tables:intu^ndu=u^(n+1)/(n+1)+C
=u^(-2+1)/(-2+1)
=-1/u
Substitute back u=x^2-6x+10
=-1/(x^2-6x+10) -----------------------------(2)
Now let's evaluate the second integral,
int6/(x^2-6x+10)^2dx
Take the constant out,
=6int1/(x^2-6x+10)^2dx
Complete the square of the term in the denominator.
=6int1/((x-3)^2+1)^2dx
Let's apply integral substitution:u=x-3
=>du=dx
=6int1/(u^2+1^2)^2du
Now use the following from the integration tables:
int1/(a^2+-u^2)^ndu=1/(2a^2(n-1))[u/(a^2+-u^2)^(n-1)+(2n-3)int1/(a^2+-u^2)^(n-1)du]
=6{1/(2(1)^2(2-1))[u/(1^2+u^2)^(2-1)+(2(2)-3)int1/(1^2+u^2)^(2-1)du]}
=6{1/2[u/(1+u^2)+int1/(1^2+u^2)du]}
Now from the integration table:int1/(a^2+u^2)du=1/aarctan(u/a)+C
=6{1/2[u/(1+u^2)+arctan(u/1)]}
=(3u)/(1+u^2)+3arctan(u)
Substitute back u=x-3
=(3(x-3))/(1+(x-3)^2)+3arctan(x-3)
=(3x-9)/(1+x^2-6x+9)+3arctan(x-3)
=(3x-9)/(x^2-6x+10)+3arctan(x-3) -------------------------(3)
Plug back the results of the integrals 2 and 3 in 1
int1/(x^2-6x+10)^2dx=1/2[-1/(x^2-6x+10)+(3x-9)/(x^2-6x+10)+3arctan(x-3)]
=1/2[(3x-9-1)/(x^2-6x+10)+3arctan(x-3)]
=1/2[(3x-10)/(x^2-6x+10)+3arctan(x-3)]
=(3x-10)/(2(x^2-6x+10))+3/2arctan(x-3)
Add a constant C to the solution,
=(3x-10)/(2(x^2-6x+10))+3/2arctan(x-3)+C