Evaluate: ∫(2x^2+1)/(x^2(x^2+4))dx

Let I = ∫(2x^2+1)/(x^2(x^2+4))dx
Perform partial fraction decomposition:
∫(2x^2+1)/(x^2(x^2+4)) = (Ax+B)/x^2 + (Cx+D)/(x^2+4)
(2x^2+1)/(x^2(x^2+4)) = (x^2(Cx+D) + x(x^2+4)A + (x^2+4)B)/(x^2(x^2+4))
After simplifying, we get
Coefficients near like terms should be equal, so the following system is obtained:
A+C=0, B+D=2, 4A=0, 4B=1.
Solve it, then A = 0, B=1/4, C = 0, D=7/4
2x^2+1 = x^3(A+C) + x^2(B+D) + 4xA + 4B
Therefore,
(2x^2+1)/(x^2(x^2+4)) = (1/4)/x^2 + (7/4)/(x^2+4)
Integrating separately, we get
= 1/4∫1/x^2 dx + 7/4∫1/(x^2+4)
We know that,
∫1/(x^2+1) = arctan(x)
= 1/4(-1/x) + 7/4 * (1/2)arctan(x/2) + C
= -1/(4x) + 7/8arctan(x/2) + C