# How do you integrate #int (x^2) / (x^2 -3x +2) # using partial fractions?

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To integrate the given function using partial fractions, we first factor the denominator: ( x^2 - 3x + 2 = (x - 1)(x - 2) ) Then, we express the given fraction as a sum of two partial fractions: ( \frac{{x^2}}{{x^2 - 3x + 2}} = \frac{{A}}{{x - 1}} + \frac{{B}}{{x - 2}} ) Multiplying both sides by ( x^2 - 3x + 2 ): ( x^2 = A(x - 2) + B(x - 1) ) Expanding and equating coefficients: ( x^2 = Ax - 2A + Bx - B ) ( x^2 = (A + B)x - (2A + B) ) Comparing coefficients: ( A + B = 1 ) ( -2A - B = 0 ) Solving these simultaneous equations, we find: ( A = -1 ) and ( B = 2 ) So, the partial fraction decomposition is: ( \frac{{x^2}}{{x^2 - 3x + 2}} = \frac{{-1}}{{x - 1}} + \frac{{2}}{{x - 2}} ) Now, we integrate each term separately: ( \int \frac{{x^2}}{{x^2 - 3x + 2}} , dx = \int \frac{{-1}}{{x - 1}} , dx + \int \frac{{2}}{{x - 2}} , dx ) ( = -\ln|x - 1| + 2\ln|x - 2| + C ) Where ( C ) is the constant of integration.

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