Text Compute ∫351(x−1)(x−2) dx\displaystyle\int_3^5 \frac{1}{(x-1)(x-2)}\,dx∫35(x−1)(x−2)1dx using partial fractions. Solution 1(x−1)(x−2)=−1x−1+1x−2\dfrac{1}{(x-1)(x-2)}=\dfrac{-1}{x-1}+\dfrac{1}{x-2}(x−1)(x−2)1=x−1−1+x−21 (from A=−1A=-1A=−1, B=1B=1B=1). ∫35(1x−2−1x−1)dx=[ln∣x−2x−1∣]35=ln32≈0.405\displaystyle\int_3^5\Big(\tfrac{1}{x-2}-\tfrac{1}{x-1}\Big)dx=\Big[\ln\Big|\tfrac{x-2}{x-1}\Big|\Big]_3^5=\ln\tfrac32\approx0.405∫35(x−21−x−11)dx=[lnx−1x−2]35=ln23≈0.405. I=ln32≈0.405\boxed{I=\ln\tfrac32\approx0.405}I=ln23≈0.405