Hàm số nào dưới đây là nguyên hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaGymaaqa % amaakaaabaGaaGymaiabgUcaRiaadIhadaahaaWcbeqaaiaaikdaaa % aabeaaaaaaaa!3EC9! f\left( x \right) = \frac{1}{{\sqrt {1 + {x^2}} }}\) trên khoảng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVy0Je9yqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq % GHsislcqGHEisPcaGG7aGaey4kaSIaeyOhIukacaGLOaGaayzkaaGa % ai4paaaa!3E11! \left( { - \infty ; + \infty } \right)?\)
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Lời giải:
Báo saiVì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2D % aebbfv3ySLgzGueE0jxyaibaieYlh9vrpeun0dXdh9vqqj-hEeeu0x % Xdbba9frFj0-OqFfea0dXdd9vqaq-JfrVkFHe9pgea0dXdar-Jb9hs % 0dXdbPYxe9vr0-vr0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaO % qaamaadmaabaGaciiBaiaac6gadaqadaqaaiaadIhacqGHRaWkdaGc % aaqaaiaaigdacqGHRaWkcaWG4bWaaWbaaSqabeaacaaIYaaaaaqaba % aakiaawIcacaGLPaaaaiaawUfacaGLDbaadaahaaWcbeqaaOGamai4 % gkdiIcaacqGH9aqpdaWcaaqaamaabmaabaGaamiEaiabgUcaRmaaka % aabaGaaGymaiabgUcaRiaadIhadaahaaWcbeqaaiaaikdaaaaabeaa % aOGaayjkaiaawMcaamaaCaaaleqabaGccWaGGBOmGikaaaqaaiaadI % hacqGHRaWkdaGcaaqaaiaaigdacqGHRaWkcaWG4bWaaWbaaSqabeaa % caaIYaaaaaqabaaaaOGaeyypa0ZaaSaaaeaacaaIXaGaey4kaSYaaS % aaaeaacaWG4baabaWaaOaaaeaacaaIXaGaey4kaSIaamiEamaaCaaa % leqabaGaaGOmaaaaaeqaaaaaaOqaaiaadIhacqGHRaWkdaGcaaqaai % aaigdacqGHRaWkcaWG4bWaaWbaaSqabeaacaaIYaaaaaqabaaaaOGa % eyypa0ZaaSaaaeaacaaIXaaabaWaaOaaaeaacaaIXaGaey4kaSIaam % iEamaaCaaaleqabaGaaGOmaaaaaeqaaaaakiaac6caaaa!6B60! {\left[ {\ln \left( {x + \sqrt {1 + {x^2}} } \right)} \right]^\prime } = \frac{{{{\left( {x + \sqrt {1 + {x^2}} } \right)}^\prime }}}{{x + \sqrt {1 + {x^2}} }} = \frac{{1 + \frac{x}{{\sqrt {1 + {x^2}} }}}}{{x + \sqrt {1 + {x^2}} }} = \frac{1}{{\sqrt {1 + {x^2}} }}.\)
Nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVy0Je9yqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iGacYgacaGGUbWaaeWa % aeaacaWG4bGaey4kaSYaaOaaaeaacaaIXaGaey4kaSIaamiEamaaCa % aaleqabaGaaGOmaaaaaeqaaaGccaGLOaGaayzkaaGaey4kaSIaam4q % aaaa!453F! F\left( x \right) = \ln \left( {x + \sqrt {1 + {x^2}} } \right) + C\)