evaluate the integral: 37/x^3-1 Solution Partial fractions, like bondman says, u

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Partial fractions, like bondman says, using x³-1 = (x-1)(x²+x+1) 1/(x³-1) = A/(x-1) + (Bx+C)/(x²+x+1) 1 = A(x²+x+1) + (Bx+C)(x-1) Immediately you can see B= -A, since the x² coefft is 0 by setting x=1, you can see A = 1/3 [1 coefft] 1 = A -C => C = (A-1) = -2/3 [x coefft] 0 = A -B +C => B = 2A-1 = -1/3 1/(x³-1) = 1/3 [ 1/(x-1) - (x+2)/(x²+x+1) ] Finally: I = 1/3 I1 - 2/3 I2 with: I1 = ? dx/(x-1) = ln(x-1) + C1 and: I2 = ? (x+2)/(x²+x+1) dx = ? (x+1/2 + 3/2)/((x+1/2)² + 3/4) dx set u = (x+1/2); du = dx Completing the square: I2 = ? (u + 3/2) / (u² + (v3/2)²) du = ? u du / (u² + (v3/2)²) + 3/2 ? du / (u² + (v3/2)²) = I3 + 3/2 I4 I3 is direct integral: 1/2 ln(u² + (v3/2)²) I4 is an arctan(...) [see third link] by substitution with a² = 3/4 ; a = v3/2 I4 = 1/a arctan(u/a) = 2/v3 arctan(2u/v3) = 2/v3 arctan(2(x + 1/2)/v3)

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