`int (dx)/(x(x^2 + 4)^2)` Evaluate the integral

`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...

`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` .