If int frac{x^2}{(x-1)(x-2)(x-3)} dx = log_e | vedclass

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(B) To solve $\int \frac{x^2}{(x-1)(x-2)(x-3)} dx$,we use partial fractions. Let $\frac{x^2}{(x-1)(x-2)(x-3)} = 1 + \frac{A}{x-1} + \frac{B}{x-2} + \f

Vedclass · Accepted Answer

$C \frac{|x-1|^{1/2} |x-3|^{9/2}}{(x-2)^4}$

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(B) To solve $\int \frac{x^2}{(x-1)(x-2)(x-3)} dx$,we use partial fractions. Let $\frac{x^2}{(x-1)(x-2)(x-3)} = 1 + \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$. Using the cover-up rule: $A = \frac{1^2}{(1-2)(1-3)} = \frac{1}{(-1)(-2)} = \frac{1}{2}$. $B = \frac{2^2}{(2-1)(2-3)} = \frac{4}{(1)(-1)} = -4$. $C = \frac{3^2}{(3-1)(3-2)} = \frac{9}{(2)(1)} = \frac{9}{2}$. Thus,the integral becomes $\int (1 + \frac{1/2}{x-1} - \frac{4}{x-2} + \frac{9/2}{x-3}) dx$. $= x + \frac{1}{2} \log_e |x-1| - 4 \log_e |x-2| + \frac{9}{2} \log_e |x-3| + C'$. $= \log_e e^x + \log_e |x-1|^{1/2} - \log_e |x-2|^4 + \log_e |x-3|^{9/2} + C'$. $= \log_e \left( \frac{e^x |x-1|^{1/2} |x-3|^{9/2}}{|x-2|^4} \right) + C'$. Comparing with $\log_e f(x)$,we get $f(x) = C \cdot \frac{|x-1|^{1/2} |x-3|^{9/2}}{(x-2)^4}$.

If $\int \frac{x^2}{(x-1)(x-2)(x-3)} dx = \log_e f(x) + C$,then $f(x) =$

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