Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 35

int (dx)/(x(x^2+4)^2)
= int 1/(x(x^2+4)^2)dx
To solve, apply partial fraction decomposition. So to express the integrand as sum of proper rational expressions, set-up the equation as follows:
1/(x(x^2+4)^2) = A/x + (Bx+C)/(x^2+4) + (Dx+E)/(x^2+4)^2
Multiply both sides by the LCD.
1=A(x^2+4)^2 + (Bx+C)(x)(x^2+4) + (Dx+E)(x)
1=Ax^4+8Ax^2+16A + Bx^4+Cx^3+4Bx^2+4Cx+Dx^2+Ex
1=(A+B)x^4 + Cx^3 + (8A+4B)x^2+(4C+E)x +16A
Express the left side as polynomial with degree 4.
0x^4 + 0x^3+0x^2+0x+1 =(A+B)x^4 + Cx^3 + (8A+4B+D)x^2+(4C+E)x +16A
For the two sides to be equal, the coefficients of the polynomial should be the same. So set the coefficients of the two polynomials equal to each other.
x^4:
0 = A + B (Let this be EQ1.)
x^3:
0=C (Let this be EQ2.)
x^2:
0=8A+4B+D (Let this be EQ3.)
x:
0=4C+E (Let this be EQ4.)
Constant:
1 = 16A (Let this be EQ5.)
Notice that the value of C is already known. So let's solve for the values of A, B, D and E.
Plug-in the value of C to EQ4.
0=4(0) + E
0=E
Isolate the A in EQ5.
1=16A
1/16=A
Then, plug-in the value of A to EQ1.
0=1/16+B
-1/16=B
And, plug-in the value of A and B to EQ3.
0=8(1/16)+4(-1/16)+D
0=8/16-4/16+D
0=4/16+D
0=1/4+D
-1/4=D
So the partial fraction decomposition of the integrand is:
1/(x(x^2+4)^2) = (1/16)/x + (-1/16x+0)/(x^2+4) + (-1/4x+0)/(x^2+4)^2 = 1/(16x) -x/(16(x^2+4))-x/(4(x^2+4)^2)
Then, proceed to take the integral of it.
int1/(x(x^2+4)^2) dx
= int (1/(16x)-x/(16(x^2+4))-x/(4(x^2+4)^2))dx
= int 1/(16x)dx - int x/(16(x^2+4)) dx - int x/(4(x^2+4)^2)dx
For the second and third integral, apply u-substitution method.
u=x^2+4
du=2x dx
(du)/2=xdx
= int 1/(16x)dx - int 1/(16u)*(du)/2 int (1/(4u^2) *(du)/2
= 1/16int 1/xdx - 1/32int 1/u*du -1/8int 1/u^2du
=1/16int 1/xdx - 1/32int 1/u*du -1/8int u^(-2)du
=1/6ln|x| -1/32 ln|u| +1/(8u)+C
And, substitute back u=x^2+4 .
=1/16ln|x|-1/32ln|x^2+4| +1/(8(x^2+4))+C

Therefore, int (dx)/(x(x^2+4)^2)=1/16ln|x|-1/32ln|x^2+4| +1/(8(x^2+4))+C .