If int frac{dx}{(x+1)(x-2)(x-3)}=frac{1}{k} l | vedclass

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(D) We use partial fractions to decompose the integrand: $\frac{1}{(x+1)(x-2)(x-3)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x-3}$.Equating the nume

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$12$

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(D) We use partial fractions to decompose the integrand: $\frac{1}{(x+1)(x-2)(x-3)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x-3}$.Equating the numerators: $1 = A(x-2)(x-3) + B(x+1)(x-3) + C(x+1)(x-2)$.For $x = -1$: $1 = A(-3)(-4) \Rightarrow A = \frac{1}{12}$.For $x = 2$: $1 = B(3)(-1) \Rightarrow B = -\frac{1}{3}$.For $x = 3$: $1 = C(4)(1) \Rightarrow C = \frac{1}{4}$.Integrating term by term: $I = \int \left( \frac{1/12}{x+1} - \frac{1/3}{x-2} + \frac{1/4}{x-3} \right) dx = \frac{1}{12} \ln|x+1| - \frac{1}{3} \ln|x-2| + \frac{1}{4} \ln|x-3| + c$.To match the form $\frac{1}{k} \ln \left\{ \frac{|x-3|^3|x+1|}{(x-2)^4} \right\}$, we factor out $\frac{1}{12}$:$I = \frac{1}{12} \left( \ln|x+1| - 4\ln|x-2| + 3\ln|x-3| \right) + c = \frac{1}{12} \ln \left\{ \frac{|x+1||x-3|^3}{|x-2|^4} \right\} + c$.Comparing this with the given expression, we find $k = 12$.

If $\int \frac{dx}{(x+1)(x-2)(x-3)}=\frac{1}{k} \log_e \left\{ \frac{|x-3|^3|x+1|}{(x-2)^4} \right\}+c$, then the value of $k$ is

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