Partial fraction question help

What is the integral of (5x^3 + 8x^2 + x + 2) / (x^2 (2x^2 + 1))?

I know I have to separate them into 4 partial fractions of A/x + B/x + ...

But I don't know how to separate the fraction for 2x^2 + 1. Anyone can help? Thank you.

You can't separate it (2x^2 + 1 that is)

A/x + B/x^2 + (Cx + D) / (2x^2 + 1)
A * x * (2x^2 + 1) + B * (2x^2 + 1) + (Cx + D) * x^2 = 5x^3 + 8x^2 + x + 2
A * (2x^3 + x) + B * (2x^2 + 1) + (Cx + D) * x^2 = 5x^3 + 8x^2 + x + 2
2Ax^3 + Ax + 2Bx^2 + B + Cx^3 + Dx^2 = 5x^3 + 8x^2 + x + 2

2Ax^3 + Cx^3 = 5x^3
2Bx^2 + Dx^2 = 8x^2
Ax = x
B = 2

Ax = x
A = 1
B = 2

2A + C = 5
2B + D = 8

2 + C = 5
C = 3

4 + D = 8
D = 4

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

We can integrate the first 2 terms easily enough

ln|x| - 2/x

The integral of 3x / (2x^2 + 1) requires u-substitution

u = 2x^2 + 1
du = 4x * dx

3x * dx / (2x^2 + 1) =>
(3/4) * du / u

Integrate:
(3/4) * ln|u| =>
(3/4) * ln|2x^2 + 1|


Integrating 4 * dx / (2x^2 + 1) requires trig-substitution

x = (sqrt(2)/2) * tan(t)
dx = (sqrt(2)/2) * sec(t)^2 * dt

4 * (sqrt(2)/2) * sec(t)^2 * dt / (2 * (1/2) * tan(t)^2 + 1) =>
2 * sqrt(2) * sec(t)^2 * dt / (1 + tan(t)^2) =>
2 * sqrt(2) * sec(t)^2 * dt / sec(t)^2 =>
2 * sqrt(2) * dt
Integrate
2 * sqrt(2) * t

Solve for t

x = (sqrt(2)/2) * tan(t)
sqrt(2) * x = tan(t)
arctan(sqrt(2) * x) = t


So we get:

ln|x| - 1/x + (3/4) * ln|1 + 2x^2| + arctan(sqrt(2) * x) + C

5x³ + 8x² + x + 2 = A/x + B/x² + (Cx + D)/(2x² + 1)

A = 1
B = 2
C = 3
D = 4