Biết F(x) là nguyên hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqaMndvLHfij5gC1rhimfMBNvxyNvgaMXfBLzgDOa % cEGWLCPDgA0LspCzMCHn2EX03E41sm9bWexLMBbXgBcf2CPn2qVrwz % qf2zLnharuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPv % MCG4uz3bqee0evGueE0jxyaibaieYlNi-xH8yiVC0xbbL8F4rqqrFf % peea0xe9Lq-Jc9vqaqpepm0xbbG8FasPYRqj0-yi0dXdbba9pGe9xq % -JbbG8A8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaa % keaaqaaaaaaaaaWdbiaadAgadaqadaWdaeaapeGaamiEaaGaayjkai % aawMcaaiabg2da9maalaaapaqaa8qacaaIXaaapaqaa8qacaWG4bGa % eyOeI0IaaGymaaaaaaa!5087! f\left( x \right) = \frac{1}{{x - 1}}\) và F(2) = 1. Khi đó F(3) bằng

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maapeaabaGaamOzamaa % bmaabaGaamiEaaGaayjkaiaawMcaaiaabsgacaWG4baaleqabeqdcq % GHRiI8aOGaeyypa0Zaa8qaaeaadaWcaaqaaiaaigdaaeaacaWG4bGa % eyOeI0IaaGymaaaacaqGKbGaamiEaaWcbeqab0Gaey4kIipakiabg2 % da9iGacYgacaGGUbWaaqWaaeaacaWG4bGaeyOeI0IaaGymaaGaay5b % SlaawIa7aiabgUcaRiaadoeaaaa!545F! F\left( x \right) = \int {f\left( x \right){\rm{d}}x} = \int {\frac{1}{{x - 1}}{\rm{d}}x} = \ln \left| {x - 1} \right| + C\)

Nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaabm % aabaGaaGOmaaGaayjkaiaawMcaaiabg2da9iaaigdacqGHuhY2ciGG % SbGaaiOBamaaemaabaGaaGOmaiabgkHiTiaaigdaaiaawEa7caGLiW % oacqGHRaWkcaWGdbGaeyypa0JaaGymaiabgsDiBlaadoeacqGH9aqp % caaIXaGaeyO0H4TaamOramaabmaabaGaamiEaaGaayjkaiaawMcaai % abg2da9iGacYgacaGGUbWaaqWaaeaacaWG4bGaeyOeI0IaaGymaaGa % ay5bSlaawIa7aiabgUcaRiaaigdaaaa!5CD6! F\left( 2 \right) = 1 \Leftrightarrow \ln \left| {2 - 1} \right| + C = 1 \Leftrightarrow C = 1 \Rightarrow F\left( x \right) = \ln \left| {x - 1} \right| + 1\)

Do đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaabm % aabaGaaG4maaGaayjkaiaawMcaaiabg2da9iGacYgacaGGUbGaaGOm % aiabgUcaRiaaigdacaGGUaaaaa!3EFA! F\left( 3 \right) = \ln 2 + 1.\)