Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 28

Integrate int(x^2-2x-1)/[(x-1)^2(x^2+1)]dx
Rewrite the rational function using partial fractions.
(x^2-2x-1)/[(x-1)^2(x^2+1)]=A/(x-1)+B/(x-1)^2+(Cx+D)/(x^2+1)
x^2-2x-1=A(x-1)(x^2+1)+B(x^2+1)+(Cx+D)(x-1)^2
x^2-2x-1=A(x^3-x^2+x-1)+Bx^2+B+(Cx+D)(x^2-2x+1)

x^2-2x-1=Ax^3-Ax^2-Ax-A+Bx^2+B
+ Cx^3-2Cx^2+Cx+Dx^2-2Dx+D

x^2-2x-1=(A+C)x^3+(-A+B-2C+D)x^2
+(A+C-2D)x+(-A+B+D)

Equate coefficients and solve for A, B, C, and D.
0=A+C
A=-C
-2=A+C-2D
-2=-C+C-2D
-2=-2D
D=1

1=-A+B-2C+D
1=C+B-2C+1
0=-1C+B
B=C

-1=-A+B+D
-1=C+B+1
-2=B+B
-2=2B
B=-1

C=-1

A=1

int(x^2-2x-1)/[(x-1)^2(x^2+1)]dx
=int[1/(x-1)]dx-int[1/(x-1)^2]dx+int[(-1x+1)/(x^2+1)]dx
=int[1/(x-1)]dx-int[1/(x-1)^2]dx+int[-x/(x^2+1)]dx+int[1/(x^2+1)]dx

The first integral follows the pattern int(du)/u=ln|u|+C

int[1/(x-1)]dx=ln|x-1|+C

Integrate the second integral using u-substitution.
Let u=x-1
(du)/(dx)=1
du=dx
-int1/(x-1)^2dx
=-intu^-2du
=1/u+C
1/(x-1)+C

Integrate the third integral using u-subsitution.
Let u=x^2+1

(du)/(dx)=2x
(dx)=(du)/(2x)
-intx/(x^2+1)dx
=-int(x)/(u)*(du)/(2x)
=-1/2ln|u|+C
=-1/2ln|x^2+1|+C

The fourth integral matches the pattern
intdx/(x^2+a^2)=(1/a)tan^-1(x/a)+C
int1/(x^2+1)dx
=tan^-1(x)+C

The final answer is:
ln|x-1|+1/(x-1)-1/2ln|x^2+1|+tan^-1(x)+C