Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 23

Integrated int10/[(x-1)(x^2+9)]dx
Solve for the variables A, B, and, C using the method of partial fractions.
10/[(x-1)(x^2+9)]=A/(x-1)+(Bx+C)/(x^2+9)
10=A(x^2+9)+(Bx+C)(x-1)
10=Ax^2+9A+Bx^2+Cx-Bx-C
10=(A+B)x^2+(C-B)x+(9A-C)
Equate coefficients and solve for A, B, and C.
0=A+B
A=-B

0=C-B
0=C+A

10=9A-C
0=A+C
10=10A
A=1
C=-1
B=-1

int10/[(x-1)(x^2+9)]dx=int[1/(x-1)+(-1x-1)/(x^2+9)]dx
=int1/(x-1)dx-intx/(x^2+9)dx-int1/(x^2+9)dx
The first integral matches the formint(du)/u=ln|u|+C
int1/(x-1)=ln|x-1|+C

Integrate the second integral using u-substitution.
Let u=x^2+9
(du)/dx=2x
dx=(du)/(2x)
-intx/(x^2+1)dx=-x/u*(du)/(2x)=-1/2ln|u|+C=-1/2ln|x^2+9|+C

The third integral matches the form int(dx)/(x^2+a^2)=1/atan^-1(x/a)+C
=-int1/(x^2+9)dx=-1/3tan^-1(x/3)+C


=int1/(x-1)dx-intx/(x^2+9)dx-int1/(x^2+9)dx
=ln|x-1|-1/2ln(x^2+9)-1/3tan^-1(x/3)+C

The final answer is:
=ln|x-1|-1/2ln(x^2+9)-1/3tan^-1(x/3)+C