I keep coming up one term short in my answer.  Will someone please integrate this problem using the partial fraction method, clearly showing the steps?

∫((1-x+2x^2-x^3)/(x(x^2+1)^2) dx Integrate by partial fractions.

(1-x+2x^2-x^3)/(x(x^2+1)^2 = A/x +(Bx^3+Cx^2+Dx+E)/(x^2+1)^2

cross-multiply by x.(x^2+1)^2 to give,

(1-x+2x^2-x^3) = A(x^2+1)^2 +(Bx^3+Cx^2+Dx+E).x

letting x = -2, -1, 0, 1, 2,

19 = 25A + 16B - 8C + 4D - 2E
5 = 4A + B - C + D - E
1 = A
1 = 4A + B + C + D + E
-1 = 25A + 16B + 8C + 4D + 2E

substituting for A = 1,

-6 = 16B - 8C + 4D - 2E  --------- ------ (1)
1 = B - C + D - E   ------------------------ (2)
-3 = B + C + D + E     -------------------- (3)
-26 = 16B + 8C + 4D + 2E  ------------- (4)

from (2) E = B - C + D - 1. Substitute for this into (1), (3) and (4), to give

-8 = 14B - 6C + 2D  ---------------------- (5)
-2 = 2B + 2D       -------------------------- (6)
-24 = 18B + 6C + 6D   ------------------- (7)

from(5), 2D = -8 - 14B + 6C. Substitute for this into (6) and (7).

6 = -12B + 6C       ------------------------- (6)
0 = -24B + 24C   --------------------------- (7)

From (7), B = C. Substituting for this into (6),

6 = -6C

C = -1

Further back-substitution gives D = 0 and E = -1.

The partial fraction coefficients are: A = 1, B = -1, C = -1, D = 0, E = -1.

The integral now becomes,

∫((1-x+2x^2-x^3)/(x(x^2+1)^2) dx = ∫ 1/x - (x^3 + x^2 + 1)/(x^2+1)^2 dx

= ∫ 1/x - (x^3 + x^2 + 1)/(x^4+2x^2+1) dx

= ln(x) - ∫ {(4x^3 + 4x) - 4x + 4x^2 + 4}/ 4(x^4 + 2x^2 + 1) dx

= ln(x) - ∫ (4x^3 + 4x)/4(x^4 + 2x^2 + 1) - (4x - 4x^2 - 4)/ 4(x^4 + 2x^2 + 1) dx

= ln(x) -(1/4)ln(x^2+1)^2 - ∫ (x^2 - x + 1)/(x^2+1)^2 dx

= ln(x) -(1/2)ln(x^2+1) - ∫ (x^2 + 1)/(x^2+1)^2 dx + ∫ (x)/(x^2+1)^2 dx

= ln(x) -(1/2)ln(x^2+1) - ∫ (1)/(x^2+1) dx - (1/2)/(x^2+1)

= ln(x) -(1/2)ln(x^2+1) - arctan(x) - (1/2)/(x^2+1)

answered Dec 1, 2013 by Level 11 User (76,040 points)