int frac{x + 2}{x^2 - 7x + 12} dx is equal to | vedclass

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Question

(A) Step $1$: Factor the denominator: $x^2 - 7x + 12 = (x - 3)(x - 4)$.Step $2$: Use partial fractions: $\frac{x + 2}{(x - 3)(x - 4)} = \frac{A}{x - 3

Vedclass · Accepted Answer

$-5 \log |x - 3| + 6 \log |x - 4| + c$

Vedclass · Answer

(A) Step $1$: Factor the denominator: $x^2 - 7x + 12 = (x - 3)(x - 4)$.Step $2$: Use partial fractions: $\frac{x + 2}{(x - 3)(x - 4)} = \frac{A}{x - 3} + \frac{B}{x - 4}$.Step $3$: Solve for $A$ and $B$: $x + 2 = A(x - 4) + B(x - 3)$.For $x = 3$: $3 + 2 = A(3 - 4) \implies 5 = -A \implies A = -5$.For $x = 4$: $4 + 2 = B(4 - 3) \implies 6 = B \implies B = 6$.Step $4$: Integrate: $\int (\frac{-5}{x - 3} + \frac{6}{x - 4}) dx = -5 \log |x - 3| + 6 \log |x - 4| + c$.

$\int \frac{x + 2}{x^2 - 7x + 12} dx$ is equal to

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