If the function f(x) = int frac{x + 3}{x^2 - | vedclass

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(C) Given $f(x) = \int \frac{x + 3}{x^2 - 9x + 20} dx$.By the Fundamental Theorem of Calculus, $f'(x) = \frac{x + 3}{x^2 - 9x + 20}$.Factor the denomi

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decreases on $(-\infty, 4) \cup (5, \infty)$

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(C) Given $f(x) = \int \frac{x + 3}{x^2 - 9x + 20} dx$.By the Fundamental Theorem of Calculus, $f'(x) = \frac{x + 3}{x^2 - 9x + 20}$.Factor the denominator: $x^2 - 9x + 20 = (x - 4)(x - 5)$.So, $f'(x) = \frac{x + 3}{(x - 4)(x - 5)}$.To find where $f(x)$ increases or decreases, we analyze the sign of $f'(x)$ using the critical points $x = -3, 4, 5$.Intervals: $(-\infty, -3)$, $(-3, 4)$, $(4, 5)$, $(5, \infty)$.For $x \in (-\infty, -3)$, $f'(x) = \frac{(-)}{(-)(-)} = (-)$, so $f(x)$ decreases.For $x \in (-3, 4)$, $f'(x) = \frac{(+)}{(-)(-)} = (+)$, so $f(x)$ increases.For $x \in (4, 5)$, $f'(x) = \frac{(+)}{(+)(-)} = (-)$, so $f(x)$ decreases.For $x \in (5, \infty)$, $f'(x) = \frac{(+)}{(+)(+)} = (+)$, so $f(x)$ increases.Thus, $f(x)$ decreases on $(-\infty, -3] \cup (4, 5)$.

If the function $f(x) = \int \frac{x + 3}{x^2 - 9x + 20} dx$, then $f(x)$ is

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