Nếu gọi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqi-u0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9maapeaabaWaaSaaaeaacaWGKbGaaeiEaaqaamaakaaabaGaaGOm % aiaabIhacqGHsislcaaIXaaaleqaaOGaey4kaSIaaGinaaaaaSqabe % qaniabgUIiYdaaaa!40EB! I = \int {\frac{{d{\rm{x}}}}{{\sqrt {2{\rm{x}} - 1} + 4}}} \), thì khẳng định nào sau đây là đúng?
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqi-u0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaada % GcaaqaaiaaikdacaqG4bGaeyOeI0IaaGymaaWcbeaakiabgkHiTiaa % isdaciGGSbGaaiOBamaabmaabaWaaOaaaeaacaaIYaGaaeiEaiabgk % HiTiaaigdaaSqabaGccqGHRaWkcaaI0aaacaGLOaGaayzkaaGaey4k % aSIaam4qaaGaayjkaiaawMcaaiaacEcacqGH9aqpdaWcaaqaaiaaig % daaeaadaGcaaqaaiaaikdacaWG4bGaeyOeI0IaaGymaaWcbeaaaaGc % cqGHsislcaaI0aWaaSaaaeaacaaIXaaabaWaaOaaaeaacaaIYaGaam % iEaiabgkHiTiaaigdaaSqabaGccqGHRaWkcaaI0aaaaiaac6cadaWc % aaqaaiaaigdaaeaadaGcaaqaaiaaikdacaWG4bGaeyOeI0IaaGymaa % Wcbeaaaaaaaa!598D! \left( {\sqrt {2{\rm{x}} - 1} - 4\ln \left( {\sqrt {2{\rm{x}} - 1} + 4} \right) + C} \right)' = \frac{1}{{\sqrt {2x - 1} }} - 4\frac{1}{{\sqrt {2x - 1} + 4}}.\frac{1}{{\sqrt {2x - 1} }}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqi-u0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaadaGcaaqaaiaaikdacaWG4bGaeyOeI0IaaGymaaWcbeaakiab % gUcaRiaaisdacqGHsislcaaI0aaabaWaaeWaaeaadaGcaaqaaiaaik % dacaWG4bGaeyOeI0IaaGymaaWcbeaakiabgUcaRiaaisdaaiaawIca % caGLPaaadaGcaaqaaiaaikdacaWG4bGaeyOeI0IaaGymaaWcbeaaaa % GccqGH9aqpdaWcaaqaaiaaigdaaeaadaGcaaqaaiaaikdacaWG4bGa % eyOeI0IaaGymaaWcbeaakiabgUcaRiaaisdaaaaaaa!4F16! = \frac{{\sqrt {2x - 1} + 4 - 4}}{{\left( {\sqrt {2x - 1} + 4} \right)\sqrt {2x - 1} }} = \frac{1}{{\sqrt {2x - 1} + 4}}\)