How to integrate (x^4+3x^2-4x+5)/ (x-1)^2(x^2+1)

Partial Fraction

Please be specific if possible.

I don't know if there's a better alternative way but it is possible and long by partial fraction

Note that your numerator and denominator are both of degree 4, which is not a proper fraction, so we need to use long division

i.e. (x^4+3x^2-4x+5) / (x^4-2x^3+2x^2-2x+1)
= 1 + (2x^3 + x^2 - 2x +4) / (x^4 -2x^3 - + x^2 - 2x + 4)
By partial fraction
(2x^3 + x^2 - 2x +4)/(x-1)^2(x^2+1) = A / (x-1) + B / (x-1)^2 + (Cx+D)/(x^2 +1)

A(x-1)(x^2 + 1) + B(x^2 + 1) + (Cx+D)(x-1)^2 = 2x^3 + x^2 - 2x +4

(A+C)x^3 + (D+B-2C-A)x^2 + (A+C-2D)x + (D-A) = 2x^3 + x^2 - 2x +4

Comparing the coefficients you have
A + C = 2
D + B - 2C - A = 1
A + C + 2D = -2
D - A = 4
Solving for A, B, C and D you get
A = -6, B = 13, C = 8, D = -2
So
(x^4+3x^2-4x+5) / (x^4-2x^3+2x^2-2x+1)
= 1 + -6 / (x-1) + 13 / (x-1)^2 + (8x -2) / (x^2+1)
= 1 + -6/(x-1) + 13/(x-1)^2 + 8x/(x^2+1) - 2/(x^2+1)
Integrate and you get
x - 6In(abs[x-1]) - 13 / (x-1) + 4In(abs[x^2+1]) - 2arctan(x)

Hello,

note: the numerator and the denominator are of the same order, so let's first expand the denominator an then long divide the numerator by the denominator:

(x^4 + 3x² - 4x + 5) /[(x - 1)²(x² + 1)] =

(x^4 + 3x² - 4x + 5) /[(x² - 2x + 1)(x² + 1)] =

(x^4 + 3x² - 4x + 5) /(x^4 - 2x³ + x² + x² - 2x + 1) =

(x^4 + 3x² - 4x + 5) /(x^4 - 2x³ + 2x² - 2x + 1)


..x^4 + 0x³ + 3x². - 4x. + 5 | x^4 - 2x³ + 2x² - 2x + 1
- x^4 + 2x³. - 2x² + 2x.. - 1 | 1
---------------------------------------
..0....+ 2x³ + x².. - 2x.. + 4

yielding:

(x^4 + 3x² - 4x + 5) /(x^4 - 2x³ + 2x² - 2x + 1) = 1 + [(2x³ + x² - 2x + 4) /(x^4 - 2x³ +
2x² - 2x + 1)]

thus the integral becomes: