int frac{2 x^3-4 x^2-x-3}{x^2-2 x-3} d x= | vedclass

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(D) First, perform polynomial long division of the numerator by the denominator: $2x^3 - 4x^2 - x - 3 = (2x)(x^2 - 2x - 3) + (5x - 3)$. Thus, the inte

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$x^2+2 \log |x+1|+3 \log |x-3|+c$

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(D) First, perform polynomial long division of the numerator by the denominator: $2x^3 - 4x^2 - x - 3 = (2x)(x^2 - 2x - 3) + (5x - 3)$. Thus, the integrand becomes: $\frac{2x^3 - 4x^2 - x - 3}{x^2 - 2x - 3} = 2x + \frac{5x - 3}{(x+1)(x-3)}$. Now, use partial fraction decomposition for $\frac{5x - 3}{(x+1)(x-3)}$: $\frac{5x - 3}{(x+1)(x-3)} = \frac{A}{x+1} + \frac{B}{x-3}$. $5x - 3 = A(x-3) + B(x+1)$. Setting $x = -1$, we get $-8 = A(-4) \implies A = 2$. Setting $x = 3$, we get $12 = B(4) \implies B = 3$. Therefore, the integral is: $I = \int (2x + \frac{2}{x+1} + \frac{3}{x-3}) dx = x^2 + 2 \log |x+1| + 3 \log |x-3| + c$.

$\int \frac{2 x^3-4 x^2-x-3}{x^2-2 x-3} d x=$

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