Integral of (x^3+x+2)/(x^4 +2x^2 +1) using Partial Fractions

The problem with this decomposition is the quadratic nature of the perfect square trinomial in the denominator (kinda the most difficult type of partial fraction you may be asked), but if you play the rules it's straight forward. Notice that for the highest degree in the denominator you need one degree less in the numerator. Moreover, for each repeated factor you need that many terms in the decomposition. So we have a quadratic factoring of a perfect square trinomial, so we have double trouble, but here it goes
[ x^3 + x + 2] / [ x^4 + 2 x^2 + 1] = [ x^3 + x + 2] / [ (x^2+1)^2 ]
The partial fraction decomposition
[ x^3 + x + 2] / [ (x^2+1)^2 ] = [ A x + B] / [ x^2 + 1] + [C x + D] / [ (x^2+1)^2 ]
So clearing the fractions we have
[ x^3 + x + 2] = [ A x + B](x^2+1) + [C x + D]
You may solve this system a number of ways, but I chose to equation coefficients, after multiplying the binomials (Ax+B)(x^2+1) and equating coefficients
x^3 : 1 = A
x^2 : 0 = B
x^1 : 1 = A + C
x^0 : 2 = B + D
Solving the system we have A = 1, B = 0, C = 0, and D = 1. Rewriting the rational expression we have
[ x^3 + x + 2] / [ (x^2+1)^2 ] = x / [x^2 + 1] + 2 / [ (x^2+1)^2 ]
Integrating the above, the first term is a simple u-substitution and the second in a trig substitution. Taking the first term you have
u = x^2 + 1 and du = 2 x dx
So,
int( 1/2 * du/u, u) = 1/2 ln|u| = 1/2 ln(x^2+1) + C
Taking the second term we have
x = tan(v) and dx = sec^2 (v) dv