Integrate the rational function: frac{3x-1}{( | vedclass

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Question

(A) Let $\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$.$3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$.For $x=1$: $3(

Vedclass · Accepted Answer

$\log |x-1| - 5\log |x-2| + 4\log |x-3| + C$

Vedclass · Answer

(A) Let $\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$.$3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$.For $x=1$: $3(1)-1 = A(1-2)(1-3) \Rightarrow 2 = A(-1)(-2) \Rightarrow 2A = 2 \Rightarrow A=1$.For $x=2$: $3(2)-1 = B(2-1)(2-3) \Rightarrow 5 = B(1)(-1) \Rightarrow -B = 5 \Rightarrow B=-5$.For $x=3$: $3(3)-1 = C(3-1)(3-2) \Rightarrow 8 = C(2)(1) \Rightarrow 2C = 8 \Rightarrow C=4$.Thus,$\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{1}{x-1} - \frac{5}{x-2} + \frac{4}{x-3}$.Integrating both sides with respect to $x$:$\int \frac{3x-1}{(x-1)(x-2)(x-3)} dx = \int \frac{1}{x-1} dx - 5 \int \frac{1}{x-2} dx + 4 \int \frac{1}{x-3} dx$.$= \log |x-1| - 5 \log |x-2| + 4 \log |x-3| + C$.

Integrate the rational function: $\frac{3x-1}{(x-1)(x-2)(x-3)}$

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