EXAMPLE 2 Evaluate SOLUTION Since the degree of the numerator is less than the degree of the denominator, we don't need

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Example 2 Evaluate Solution Since The Degree Of The Numerator Is Less Than The Degree Of The Denominator We Don T Need 1 (87.18 KiB) Viewed 49 times

EXAMPLE 2 Evaluate SOLUTION Since the degree of the numerator is less than the degree of the denominator, we don't need to divide. We factor the denominator as 2x³ + 3x²2x = x(2x² + 3x - 2) = x(2x - 1)(x + 2). Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand has the formt 4x² + 7x-1 dx. 2x³ + 3x² - 2x 4x² + 7x - 1 x(2x - 1)(x + 2) To determine the values of A, B, and C, we multiply both sides of this equation by the product of the denominators, x(2x - 1)(x + 2), obtaining 4x² + 7x1 = A 2x - 1 )(x + 2) + Bx(x + 2) + Cx(2x - 1). Expanding the right side of the equation above and writing it in the standard form for polynomials, we get 4x² + 7x1 = (2A + B + 2C)x² + The polynomials in the equation above are identical, so their coefficients must be equal. The coefficient of x² on the right side, 2A + B + 2C, must equal the coefficient of x² on the left side--namely, 4. Likewise, the coefficients of x are equal and the constant terms are equal. This gives the following system of equations for A, B, and C. 2A + B + 2C = 3A + 2B - C = -24 Solving, we get A = appropriate.) 1/2 = -1 B = 4x² + 7x-1 dx = 2x3 + 3x²2x /12/² and C= - 2A. + K and so we have the following. (Remember to use absolute values where ) 2 X ² = 1 + (1 )x+ 2] dx In integrating the middle term we have made the mental substitution u = 2x 1, which gives du = 2 dx and dx = =10 du.