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Integration with partial fractions? Help?
int 1/(x^2)*(x-1)^2 dx
Do I distribute the (x-1)^2 with x^2 ?? Or do I go right ahead and do
A/x^2 + B/x-1 + C/(x-1)^2 and proceed???
1 個解答
- 1 十年前最愛解答
A / x + B / x^2 + C / (x - 1) + D / (x - 1)^2 = 1 / (x^2 * (x - 1)^2)
A * (x) * (x - 1)^2 + B * (x - 1)^2 + C * (x^2) * (x - 1) + D * (x^2) = 1
A * x * (x^2 - 2x + 1) + B * (x^2 - 2x + 1) + C * (x^3 - x^2) + Dx^2 = 1
(Ax + B) * (x^2 - 2x + 1) + Cx^3 - Cx^2 + Dx^2 = 1
Ax^3 - 2Ax^2 + Ax + Bx^2 - 2Bx + B + Cx^3 - Cx^2 + Dx^2 = 1
Ax^3 + Cx^3 = 0x^2
-2Ax^2 + Bx^2 - Cx^2 + Dx^2 = 0x^2
Ax - 2Bx = 0
B = 1
A + C = 0
-2A + B - C + D = 0
A - 2B = 0
B = 1
A - 2 = 0
A = 2
A + C = 0
2 + C = 0
C = -2
-2A + B - C + D = 0
-2 * (2) + (1) - (-2) + D = 0
-4 + 1 + 2 + D = 0
-1 + D = 0
D = 1
2 / x + 1 / x^2 - 2 / (x - 1) + 1 / (x - 1)^2
Integrate
2 * ln(x) - x^(-1) - 2 * ln(x - 1) - (x - 1)^(-1) + C
2 * (ln(x) - ln(x - 1)) - (1/x + 1/(x - 1)) + C
2 * ln(x / (x - 1)) - ((x - 1 + x) / (x * (x - 1))) + C
2 * ln(x / (x - 1)) - ((2x - 1) / (x^2 - x)) + C