If int frac{3x + 7}{x^2 - 3x + 2} dx = m log | vedclass

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(C) Step $1$: Factorize the denominator: $x^2 - 3x + 2 = (x - 2)(x - 1)$.Step $2$: Use partial fractions: $\frac{3x + 7}{(x - 2)(x - 1)} = \frac{A}{x

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$3$

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(C) Step $1$: Factorize the denominator: $x^2 - 3x + 2 = (x - 2)(x - 1)$.Step $2$: Use partial fractions: $\frac{3x + 7}{(x - 2)(x - 1)} = \frac{A}{x - 2} + \frac{B}{x - 1}$.Step $3$: Solve for $A$ and $B$: $3x + 7 = A(x - 1) + B(x - 2)$.For $x = 2$: $3(2) + 7 = A(2 - 1) \implies A = 13$.For $x = 1$: $3(1) + 7 = B(1 - 2) \implies 10 = -B \implies B = -10$.Step $4$: Integrate: $\int (\frac{13}{x - 2} - \frac{10}{x - 1}) dx = 13 \log |x - 2| - 10 \log |x - 1| + c$.Step $5$: Compare with $m \log |x - 2| + n \log |x - 1| + c$, we get $m = 13$ and $n = -10$.Step $6$: Calculate $m + n = 13 + (-10) = 3$.

If $\int \frac{3x + 7}{x^2 - 3x + 2} dx = m \log |x - 2| + n \log |x - 1| + c$, where $m, n \in R$ and $c$ is an integration constant, then $m + n =$

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