int (x^2+6x+4)/(x^4+8x^2 + 16) dx Use partial fractions to find the indefinite integral

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^2at 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 .