∫ {1 + [(2x³ + x² - 2x + 4) /(x^4 - 2x³ + 2x² - 2x + 1)]} dx =
(splitting into two integrals)
∫ dx + ∫ [(2x³ + x² - 2x + 4) /(x^4 - 2x³ + 2x² - 2x + 1)] dx =
x + ∫ [(2x³ + x² - 2x + 4) /(x^4 - 2x³ + 2x² - 2x + 1)] dx (#)
let's decompose the remaining integrand into partial fractions, rewriting the denominator in factored form:
(2x³ + x² - 2x + 4) /[(x - 1)²(x² + 1)] = A/(x - 1) + B/(x - 1)² + (Cx + D)/(x² + 1)
(2x³ + x² - 2x + 4) /[(x - 1)²(x² + 1)] = [A(x - 1)(x² + 1) + B(x² + 1) + (Cx + D)(x -
1)²] /[(x - 1)²(x² + 1)]
(equating numerators)
2x³ + x² - 2x + 4 = A(x³ - x² + x - 1) + Bx² + B + (Cx + D)(x² - 2x + 1)
2x³ + x² - 2x + 4 = Ax³ - Ax² + Ax - A + Bx² + B + Cx³ - 2Cx² + Cx + Dx² - 2Dx + D
2x³ + x² - 2x + 4 = (A + C)x³ + (- A + B - 2C + D)x² + (A + C - 2D)x + (- A + B + D)
(equating coefficients)
A + C = 2
- A + B - 2C + D = 1
A + C - 2D = - 2
- A + B + D = 4
C = 2 - A
- A + B - 2(2 - A) + D = 1 → - A + B - 4 + 2A + D = 1 → A + B + D = 1 + 4
A + (2 - A) - 2D = - 2 → A + 2 - A - 2D = - 2 → - 2D = - 2 - 2 → 2D = 4
- A + B + D = 4
C = 2 - A
A + B + 2 = 5 → A = 5 - B - 2 → A = 3 - B
D = 4/2 = 2
- A + B + 2 = 4 → - A + B = 4 - 2 → - A + B = 2
C = 2 - A = 2 - (1/2) = 3/2
A = 3 - B = 3 - (5/2) = 1/2
D = 2
- (3 - B) + B = 2 → - 3 + B + B = 2 → 2B = 2 + 3 → 2B = 5 → B = 5/2
obtaining:
(2x³ + x² - 2x + 4) /[(x - 1)²(x² + 1)] = A/(x - 1) + B/(x - 1)² + (Cx + D)/(x² + 1) =
(1/2)/(x - 1) + (5/2)/(x - 1)² + [(3/2)x + 2]/(x² + 1)
thus the above (#) expression becomes:
x + ∫ [(2x³ + x² - 2x + 4) /(x^4 - 2x³ + 2x² - 2x + 1)] dx = x + ∫ { [(1/2) /(x - 1)] +
[(5/2) /(x - 1)²] + {[(3/2)x + 2] /(x² + 1)} } dx =
let's split this (pulling constants out) into:
x + (1/2) ∫ [1 /(x - 1)] dx + (5/2) ∫ [1 /(x - 1)²] dx + (3/2) ∫ [x /(x² + 1)] dx +
2 ∫ [1 /(x² + 1)] dx =
(dividing and multiplying the third integral by 2 to make the numerator the derivative of the denominator)
x + (1/2) ln |x - 1| + (5/2) ∫ (x - 1)^(- 2) d(x - 1) + (3/2)(1/2) ∫ [2x /(x² + 1)] dx +
2arctanx =
x + (1/2) ln |x - 1| + (5/2) [1/(- 2+1)] (x - 1)^(- 2+1) + (3/4) ∫ d(x² + 1) /(x² + 1) +
2arctanx =
x + (1/2) ln |x - 1| + (5/2) [1/(- 1)] (x - 1)^(- 1) + (3/4) ln (x² + 1) + 2arctanx + C =
x + (1/2) ln |x - 1| - (5/2)[1 /(x - 1)] + (3/4) ln (x² + 1) + 2arctanx + C
thus the answer is:
∫ {(x^4 + 3x² - 4x + 5) /[(x - 1)²(x² + 1)]} = x + (1/2) ln |x - 1| - {5 /[2(x - 1)]} +
(3/4) ln (x² + 1) + 2arctanx + C
I hope it's helpful..