If int frac{1}{(x-2)^5(x-1)^4} d x=sum_{r=-4} | vedclass

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(A) Let $I = \int \frac{1}{(x-2)^5(x-1)^4} dx$. We can rewrite the integrand as $I = \int \frac{1}{(x-2)^9 \left(\frac{x-1}{x-2}\right)^4} dx$. Let $t

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$\log (x-2)-\log (x-1)$

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(A) Let $I = \int \frac{1}{(x-2)^5(x-1)^4} dx$. We can rewrite the integrand as $I = \int \frac{1}{(x-2)^9 \left(\frac{x-1}{x-2}\right)^4} dx$. Let $t = \frac{x-1}{x-2}$. Then $dt = \frac{(x-2) - (x-1)}{(x-2)^2} dx = \frac{-1}{(x-2)^2} dx$,so $dx = -(x-2)^2 dt$. Since $t = \frac{x-1}{x-2} = 1 + \frac{1}{x-2}$,we have $\frac{1}{x-2} = t-1$,so $x-2 = \frac{1}{t-1}$. Substituting these into the integral: $I = \int \frac{-(t-1)^2 dt}{(t-1)^{-7} t^4} = -\int (t-1)^9 t^{-4} dt$. Expanding $(t-1)^9$ using the binomial theorem,we get terms of the form $t^k$ for $k \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$. The integral of $t^{-1}$ term results in $\log |t| = \log |\frac{x-1}{x-2}| = \log |x-1| - \log |x-2|$. Comparing this with the given form $\sum A_r (\frac{x-2}{x-1})^r + B f(x)$,we identify $f(x) = \log |\frac{x-2}{x-1}| = \log (x-2) - \log (x-1)$.

If $\int \frac{1}{(x-2)^5(x-1)^4} d x=\sum_{r=-4}^{-1} A_r\left(\frac{x-2}{x-1}\right)^r+\sum_{r=1}^3 A_r\left(\frac{x-2}{x-1}\right)^r+B f(x)$,then $f(x)=$

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