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?
asked Mar 8, 2013 in Calculus Answers by anonymous

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1 Answer

∫((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 Fermat Level 10 User (61,960 points)
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