`int (x^2+6x+4)/(x^4+8x^2+16)dx`

To solve using partial fraction method, the denominator of the integrand should be factored.

`(x^2+6x+4)/(x^4+8x^2+16) = (x^2+6x+4) / (x^2+4)^2`

If the factor in the denominator is quadratic and repeating, the partial fraction of this factor is ` (A_1x+B_1)/(ax^2+bx+c)+(A_2x+B_2)/(ax^2+bx+c)^2 + ... +(A_nx+B_n)/(ax^2+bx+C)^n` .

So expressing the integrand as sum of fractions, it becomes:

`(x^2+6x+4) / (x^2+4)^2=(Ax+B)/(x^2+4) + (Cx+D)/(x^2+4)^2`

To solve for the values of A, B, C and D, multiply both sides by the LCD.

`(x^2+4)^2 * (x^2+6x+4) / (x^2+4)^2=((Ax+B)/(x^2+4) + (Cx+D)/(x^2+4)^2) * (x^2+4)^2`

`x^2+6x+4=(Ax + B)(x^2+4) + Cx + D`

`x^2+6x + 4 = Ax^3 + 4Ax + Bx^2 + 4B + Cx + D`

At the right side, group together the terms with same power of x.

`x^2+6x+4 =Ax^3 + Bx^2 + (4Ax + Cx) + (4B + D)`

`x^2+6x+4=Ax^3 + Bx^2 + (4A+C)x + (4B + D)`

Notice that the right side has a degree of 3. So express the polynomial at the left side with a degree of 3.

`0x^3+x^2+6x+4=Ax^3 + Bx^2 + (4A+C)x + (4B + D)`

In order that the two polynomials to be equal, the coefficients and the constant should be the same.

So set the coefficient of `x^3` at the left side equal to the coefficient of `x^3` at the right side.

`0=A`

Also, set the coefficient of `x^2`at the left side equal to the coefficient of `x^2` at the right side.

`1=B`

Set the coefficient of x at the left side equal to the coefficient of x at the right side too.

`6=4A + C` (Let this be EQ1.)

And set the constant at the left side equal to the constant at the right side.

`4=4B+D` (Let this be EQ2.)

Since the values of A and B are known already, plug-in them to equation 1 and 2 to get the values of C and D.

Plug-in A=0 to EQ1 to get the value of C.

`6=4(0) +C`

`6=C`

And, plug-in B = 1 to EQ2 to get the value of D.

`4=4(1)+D`

`4=4+D`

`0=D`

So the partial fraction decomposition of the integrand is:

`int (x^2+6x+4)/(x^4+8x^2+16)dx`

= `int(x^2+6x+4) / (x^2+4)^2dx`

`=int (1/(x^2+4) + (6x)/(x^2+4)^2)dx`

Expressing it as sum of two integrals, it becomes:

`= int 1/(x^2+4)dx + int (6x)/(x^2+4)^2 dx`

`= int 1/(x^2+4)dx + 6int (x)/(x^2+4)^2 dx`

For the first integral, apply the formula `int 1/(u^2+a^2) du = 1/a tan^(-1) (u/a) + C` .

`u = x`

`du = dx`

`a=2`

For the second integral, apply the formula `int u^n du = u^(n+1)/(n+1)+C` .

`u = x^2+4`

`du = 2x dx`

So the result of each integral is:

`= int 1/(x^2+4)dx + 6int (x^2+4)^(-2) *xdx`

`= int 1/(x^2+4)dx + 3int (x^2+4)^(-2) *2xdx`

`= 1/2 tan^(-1)(x/2) + 3*(x^2+4)^(-1)/(-1)+C`

`= 1/2 tan^(-1)(x/2) - 3(x^2+4)^(-1)+C`

`= 1/2 tan^(-1)(x/2) - 3/(x^2+4)+C`

**Therefore, `int (x^2+6x+4)/(x^4+8x^2+16)dx= 1/2 tan^(-1)(x/2) - 3/(x^2+4)+C` .**