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

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Aicharya · Accepted Answer

We are tasked with evaluating the following integral: $ \int \frac{dx}{(x+1)(x-2)(x-3)} = \int \left( \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x-3} \right) dx $ We start by equating the right-hand side expression to the left-hand side: $ \frac{1}{(x+1)(x-2)(x-3)} = \frac{A(x-2)(x-3) + B(x+1)(x-3) + C(x+1)(x-2)}{(x+1)(x-2)(x-3)} $ Thus, we get the equation: $ 1 = A(x-2)(x-3) + B(x+1)(x-3) + C(x+1)(x-2) $ Now, we will solve for $A$, $B$, and $C$ by substituting convenient values for $x$. 1. **For $x = -1$:** $ 1 = A(-3)(-4) \Rightarrow A = \frac{1}{12} $ 2. **For $x = 2$:** $ 1 = B(3)(-1) \Rightarrow B = -\frac{1}{3} $ 3. **For $x = 3$:** $ 1 = C(4)(1) \Rightarrow C = \frac{1}{4} $ Thus, we have the values $A = \frac{1}{12}$, $B = -\frac{1}{3}$, and $C = \frac{1}{4}$. Now, substitute these values into the integral: $ I = \frac{1}{12} \int \frac{1}{x+1} dx - \frac{1}{3} \int \frac{1}{x-2} dx + \frac{1}{4} \int \frac{1}{x-3} dx $ Evaluating each integral gives: $ I = \frac{1}{12} \ln |x+1| - \frac{1}{3} \ln |x-2| + \frac{1}{4} \ln |x-3| + C $ Next, combine the logarithms: $ I = \frac{1}{12} \left[\ln |x+1| - \ln |(x-2)^4| + \ln |(x-3)^3| \right] + C $ Simplifying further: $ I = \frac{1}{12} \ln \left( \frac{|x+1| |(x-3)^3|}{(x-2)^4} \right) + C $ Finally, we conclude that: $ \boxed{k = 12} $

If \(\int \frac{d x}{(x+1)(x-2)(x-3)}=\frac{1}{k} \log _e\left\{\frac{| x-3|^3| x+1|}{(x-2)^4}\right\}+c\),…

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