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

I know that I cannot integrate the function as is, and need to simplify and integrate it using Integration by Partial Fractions. I'm just not sure how to go about this. Please help! :)

There's a trick to factoring x^4 + 2x^2 + 1. Let y = x^2. Then y^2 + 2y + 1 which is easily factored into (y + 1)^2. Then we have (x^2 + 1)^2.

(x^3 + x + 2)/(x^4 + 2x^2 + 1) = (Ax+B)/(x^2+1) + (Cx+D)/(x^2+1)^2

Multiply both sides by (x^2 + 1)^2.

x^3 + x + 2 = (Ax + B)(x^2 + 1) + (Cx + D)
x^3 + x + 2 = Ax^3 + Ax + Bx^2 + B + Cx + D
x^3 + x + 2 = Ax^3 + Bx^2 + (A + C) x + B + D

This forces
A = 1
B = 0
A + C = 1
B + D = 2

Thus, we have
A = 1
B = 0
C = 0
D = 2.

(x^3 + x + 2)/(x^4 + 2x^2 + 1) = x/(x^2+1) + 2/(x^2+1)^2

INT {x/(x^2+1) + 2/(x^2+1)^2} dx
INT {x/(x^2+1)} dx + INT {2/(x^2+1)^2} dx

For the first integral, let u = x^2 + 1. Then du = 2x dx or (1/2) du = x dx.

INT {x/(x^2+1)} dx = INT {(1/2) / u} du = (1/2) ln |u| = (1/2) ln (x^2 + 1)

For the second integral, let x = tan y. Then x^2 + 1 = (tan y)^2 + 1 = (sec y)^2 Also note that dx = (sec y)^2 dy

INT {2/(x^2+1)^2} dx = 2 * INT {(sec y)^2 / ((tan y)^2 + 1)^2} dy
INT {2/(x^2+1)^2} dx = 2 * INT {(sec y)^2 / ((sec y)^2)^2} dy
INT {2/(x^2+1)^2} dx = 2 * INT {(sec y)^2 / ((sec y)^4} dy
INT {2/(x^2+1)^2} dx = 2 * INT {1 / ((sec y)^2} dy
INT {2/(x^2+1)^2} dx = 2 * INT {(cos y)^2} dy

Use the formula (cos y)^2 = (1/2)(1 + cos 2y)

INT {2/(x^2+1)^2} dx = 2 * INT {(1/2)(1 + cos 2y)} dy
INT {2/(x^2+1)^2} dx = INT {1 + cos 2y} dy
INT {2/(x^2+1)^2} dx = y + (1/2) sin 2y
INT {2/(x^2+1)^2} dx = arctan x + (1/2) sin (2(arctan x))

Use the formula sin (2(arctan x)) = 2 sin (arctan x) cos (arctan x)

Since x = tan y, draw a right triangle. Label a non-right angle y. The side opposite y is of length x and the side adjacent is of length 1. This forces the hypotenuse to be of length sqrt(x^2 + 1).