What is the integral of...

using partial fractions

(x^4+1)/(x^5+4x^3)

∫ (x⁴ + 1) / (x⁵ + 4x³) dx
= ∫ (x⁴ + 1) / (x³(x² + 4)) dx

Then we are looking for something in the form:
(x⁴ + 1) / (x³(x² + 4)) = [(Ax² + Bx + C) / x³] + [(Dx + E) / (x² + 4)]

Then:
(x⁴ + 1) = (Ax² + Bx + C)(x² + 4) + (Dx +E)x³
==> x⁴ + 1 = Ax⁴ + Bx³ + 4Ax² + 4Bx + Cx² + 4C + Dx⁴ + Ex³
==> x⁴ + 1 = (A + D)x⁴ + (B + E)x³ + (4A + C)x² + (4B)x + (4C)

Set the coefficients equal to get the following system of equations:
1 = A + D
0 = B + E
0 = 4A + C
0 = 4B
1 = 4C

Solving this gives us:
C = 1/4, A = -1/16, D = 17/16, B = 0, E = 0

Thus our integral becomes:
∫ [(-(1/16)x² + (1/4)) / x³] + [(17/16)x / (x² + 4)] dx
= (-1/16) ∫ (1/x) dx + (1/4) ∫ (1/x³) dx + (17/16) ∫ x / (x² + 4) dx

Now you should be able to integrate this to get:
(-1/16)ln|x| - (1/8)(1/x²) + (17/32)ln(x² + 4) + C

the denominator can be rewritten as x^3(x^2+4)
So the partial fractions(According to rule) becomes
A/x+B/x^2+C/x^3+(Dx)/(x^2+4)...1{ note for x^3(x is a repeated first degree factor and (x^2+4) is a non repeated second degree factor, so Dx}
Now solve 1 to get numerator as
Ax^2(x^2+4) + Bx(x^2+4)+C(x^2+4)+(Dx)x^3=x^4+1
Equating coefficients of x^4 in LHS=1 and constant term =1 we get A=-1/16 B=0, C=1/4 and D=17/16
So the partial fractions are
17/16*x/(x^2+4)+1/4*1/x^3-1/16*1/x
Integrating we get 17/32log(x^2+4)-1/16log x -1/8x^(-2)+c
or 1/16{log [(x^2+4)^(17/2)]/x} -1/8x^(-2)+c where c is a constant