int frac{x , dx}{(x-1)(x-2)} = | vedclass

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(C) To evaluate the integral $\int \frac{x \, dx}{(x-1)(x-2)}$,we use partial fractions. Let $\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$. M

Vedclass · Accepted Answer

$\log \left| \frac{(x-2)^2}{x-1} \right| + c$

Vedclass · Answer

(C) To evaluate the integral $\int \frac{x \, dx}{(x-1)(x-2)}$,we use partial fractions. Let $\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$. Multiplying both sides by $(x-1)(x-2)$,we get $x = A(x-2) + B(x-1)$. Setting $x = 1$,we get $1 = A(1-2) \implies A = -1$. Setting $x = 2$,we get $2 = B(2-1) \implies B = 2$. Thus,$\int \frac{x \, dx}{(x-1)(x-2)} = \int \left( \frac{-1}{x-1} + \frac{2}{x-2} \right) dx$. Integrating term by term,we get $-\log |x-1| + 2 \log |x-2| + c$. Using the property of logarithms,$n \log m = \log m^n$,we get $\log |x-2|^2 - \log |x-1| + c = \log \left| \frac{(x-2)^2}{x-1} \right| + c$.

$\int \frac{x \, dx}{(x-1)(x-2)} =$

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