$I=\int \frac{1}{{\left(x-2\right)}^5.{\left(x-1\right)}^4}dx$

$\Rightarrow I=\int \frac{dx}{{\left(x-2\right)}^9.{\left({\displaystyle \frac{x-1}{x-2}}\right)}^4}$

Let $\frac{x-1}{x-2}=t$

$\Rightarrow \frac{\left(x-2\right).1-\left(x-1\right).1}{{\left(x-2\right)}^2}dx=dt$

$\Rightarrow \frac{-1}{{\left(x-2\right)}^2}dx=dt ...\left(1\right)$

Also, $\frac{x-1}{x-2}=t$

$\Rightarrow x-1=tx-2t$

$\Rightarrow x=\frac{2t-1}{t-1}$

$\Rightarrow x-2=\frac{1}{t-1} ...\left(2\right)$

$\therefore I=\int \frac{-dt}{{\displaystyle \frac{1}{{\left(t-1\right)}^7}}}{t}^4$

$\Rightarrow I=\int \frac{{\left(1-t\right)}^7}{t^4}dt$

$\Rightarrow I=\int \frac{1-{}^{7}c_{1}t+{}^{7}c_2{t}^2-{}^{7}c_3{t}^3+......+{}^{7}c_6.t^6-{}^{7}c_7{t}^7}{t^4}dt$

$\Rightarrow I=\int \frac{1}{t^4}dt-{}^{7}c_1\int \frac{dt}{t^3}+{}^{7}c_2\int \frac{dt}{t^2}-{}^{7}c_3\int \frac{1}{t}dt+....-{}^{7}c_7\int t^{3}dt$

$\Rightarrow I=\frac{t^{-4+1}}{-4+1}-7 \frac{t^{-3+1}}{-3+1}+{}^{7}c_2\frac{t^{-2+1}}{-2+1}-{}^{7}c_3\log \left|t\right|+...-\frac{t^4}{4}+c$

  $\Rightarrow I=\frac{-1}{3}{\left(\frac{x-1}{x-2}\right)}^{-3}+\frac{7}{2}{\left(\frac{x-1}{x-2}\right)}^{-2}-{}^{7}c_3\log \left|\frac{x-1}{x-2}\right|+.....- \frac{1}{4}{\left(\frac{x-1}{x-2}\right)}^4+c$

$\Rightarrow I=\frac{-1}{3}{\left(\frac{x-2}{x-1}\right)}^3+\frac{7}{2}{\left(\frac{x-2}{x-1}\right)}^2+{}^{7}c_2\left(\frac{x-2}{x-1}\right)+{}^{7}c_4{\left(\frac{x-2}{x-1}\right)}^{-1}+{}^{7}c_3\left[\log \left(x-2\right)-\log \left(x-1\right)\right]....-\frac{1}{4}{\left(\frac{x-2}{x-1}\right)}^{-4}+c ...\left(1\right)$

$\because$Given that

$\int \frac{1}{(x-2{)}^5(x-1{)}^4}dx=\sum _{r=-4}^{-1}{A}_r{\left(\frac{x-2}{x-1}\right)}^r+\sum _{r=1}^3{A}_r{\left(\frac{x-2}{x-1}\right)}^r+Bf\left(x\right)$

$\therefore$We get $f\left(x\right)=\log \left(x-2\right)-\log \left(x-1\right)$