Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 28

int(x^2+1)/(x^2-2x+2)^2dx
Let's evaluate the above integral by rewriting the integrand as,
int(x^2-2x+2+2x-1)/(x^2-2x+2)^2dx
=int(x^2-2x+2)/(x^2-2x+2)^2dx+int(2x-1)/(x^2-2x+2)^2dx
=int1/(x^2-2x+2)dx+int(2x-1)/(x^2-2x+2)^2dx
Now again rewrite the second integral,
=int1/(x^2-2x+2)dx+int((2x-2)+1)/(x^2-2x+2)^2dx
=int1/(x^2-2x+2)dx+int(2x-2)/(x^2-2x+2)^2dx+int1/(x^2-2x+2)^2dx
Now let's evaluate the above three integrals,
int1/(x^2-2x+2)dx=int1/((x-1)^2+1)dx
Let's use the integral substitution,
Let u=x-1,
du=dx
=int1/(1+u^2)du
The above can be evaluated using the standard integral,
int1/(x^2+a^2)dx=1/atan^(-1)(x/a)
=1/1tan^(-1)(u/1)
=tan^(-1)(u)
Substitute back u=x-1,
=tan^(-1)(x-1)
Now let's evaluate the second integral by integral substitution,
int(2x-2)/(x^2-2x+2)^2dx
Let v=(x^2-2x+2)
dv=(2x-2)dx
=int1/v^2dv
=v^(-2+1)/(-2+1)
=v^(-1)/-1
=-1/v
Substitute back v=(x^2-2x+2)
=-1/(x^2-2x+2)
Noe let's evaluate the third integral,
int1/(x^2-2x+2)^2dx
=int1/((x-1)^2+1)^2dx
Let's use the integral substitution,
Let tan(y)=x-1
sec^2(y)dy=dx
=int(sec^2(y)dy)/(tan^2(y)+1)^2
Using the identity:1+tan^2(y)=sec^2(y)
=int(sec^2(y))/(sec^2(y))^2dy
=int1/(sec^2(y))dy
=intcos^2(y)dy
Now use the identity:cos^2(y)=(1+cos(2y))/2
=int(1+cos(2y))/2dy
=int(1dy)/2+intcos(2y)/2dy
=y/2+1/2sin(2y)/2
=y/2+sin(2y)/4
Substitute back y=tan^(-1)(x-1)
=1/2arctan(x-1)+1/4sin(2arctan(x-1))
=1/2arctan(x-1)+1/4{2sin(arctan(x-1))cos(arctan(x-1))}
=1/2arctan(x-1)+1/4{2*(x-1)/sqrt(x^2-2x+2)*1/sqrt(x^2-2x+2)}
=1/2arctan(x-1)+(1/2)(x-1)/(x^2-2x+2)
:.int(x^2+1)/(x^2-2x+2)^2dx=arctan(x-1)-1/(x^2-2x+2)+1/2arctan(x-1)+(x-1)/(2(x^2-2x+2))
Add a constant C to the solution and simplify,
=3/2arctan(x-1)+(-2+x-1)/(2(x^2-2x+2))+C
=3/2arctan(x-1)+(x-3)/(2(x^2-2x+2))+C