Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 9

int (x^2+12x+12)/(x^3-4x)dx
To solve using partial fraction method, the denominator of the integrand should be factored.
(x^2+12x+12)/(x^3-4x) =(x^2+12x+12)/(x(x-2)(x+2))
Then, express it as sum of fractions.
(x^2+12x+12)/(x(x-2)(x+2)) = A/x + B/(x-2) + C/(x+2)
To determine the values of A, B and C, multiply both sides by the LCD of the fractions present.
x(x-2)(x+2)*(x^2+12x+12)/(x(x-2)(x+2)) = (A/x + B/(x-2) + C/(x+2))*x(x-2)(x+2)
x^2+12x+12=A(x-2)(x+2) +Bx(x+2)+Cx(x-2)
Then, assign values to x in which either x, x-2 or x+2 will become zero.
So, plug-in x=0 to get the value of A.
0^2+12(0)+12=A(0-2)(0+2)+B(0)(0+2)+C(0)(0-2)
0+0+12=A(-4)+B(0)+C(0)
12=-4A
-3=A
Also, plug-in x=2 to get the value of B.
2^2+12(2)+12=A(2-2)(2+2)+B(2)(2+2)+C(2)(2-2)
4+24+12=A(0)+B(8)+C(0)
40=8B
5=B
And subsitute x=-2 to get the value of C.
(-2)^2 + 12(-2)+12=A(-2-2)(-2+2)+B(-2)(-2+2)+C(-2)(-2-2)
4-24+12=A(0)+B(0)+C(8)
-8=8C
-1=C
So the partial fraction decomposition of the integral is
int (x^2+12x+12)/(x^3-4x)dx
= int (x^2+12x+12)/(x(x-2)(x+2))dx
= int(-3/x +5/(x-2)-1/(x+2))dx
Then, express it as three integrals.
= int-3/xdx + int 5/(x-2)dx - int 1/(x+2)dx
= -3int 1/xdx + 5int 1/(x-2)dx - int 1/(x+2)dx
To take the integral, apply the formula int 1/u du =ln|u|+C .
=-3ln|x| + 5ln|x-2|-ln|x+2|+C

Therefore,int (x^2+12x+12)/(x^3-4x)dx=-3ln|x| + 5ln|x-2|-ln|x+2|+C.