Với phép biến đổi \(u=\sqrt x\), tích phân \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9maapehabaWaaSaaaeaacaWGLbWaaWbaaSqabeaadaGcaaqaaiaa % dIhaaWqabaaaaaGcbaWaaOaaaeaacaWG4baaleqaaaaakiaadsgaca % WG4baaleaacaaIXaaabaGaaGinaaqdcqGHRiI8aaaa!40FB! \int\limits_1^4 {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}dx} \) trở thành
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Lời giải:
Báo saiĐặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWG1bGaeyypa0ZaaOaaa8aabaWdbiaadIhaaSqabaGccqGHshI3 % caWG1bWaaWbaaSqabeaacaaIYaaaaOGaeyypa0JaamiEaiabgkDiEl % aaikdacaWG1bGaamizaiaadwhacqGH9aqpcaWGKbGaamiEaaaa!4883! % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aacaWG1bGaeyypa0ZaaOaaaeaacaWG4baaleqaaOGaeyO0H4Taamiz % aiaadwhacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaWaaOaaaeaaca % WG4baaleqaaaaakiaadsgacaWG4bGaeyO0H4TaaGOmaiaadsgacaWG % 1bGaeyypa0ZaaSaaaeaacaaIXaaabaWaaOaaaeaacaWG4baaleqaaa % aakiaadsgacaWG4baaaa!4CDF! u = \sqrt x \Rightarrow du = \frac{1}{{2\sqrt x }}dx \Rightarrow 2du = \frac{1}{{\sqrt x }}dx\)
Đổi cận
\(\begin{array}{c|c} x & 1 \,\,\,\,\,\,\,\,\,\,4 \\ \hline u & 1\,\,\,\,\,\,\,\,\,\,2 \end{array}\)
Khí đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGjbGaeyypa0Zaa8qCaeaacaWGLbWaaWbaaSqabeaacaWG1baa % aOGaaiOlaiaaikdacaWGKbGaamyDaiabg2da9aWcbaGaaGymaaqaai % aaikdaa0Gaey4kIipakiaaikdadaWdXbqaaiaadwgadaahaaWcbeqa % aiaadwhaaaGccaWGKbGaamyDaaWcbaGaaGymaaqaaiaaikdaa0Gaey % 4kIipaaaa!4ADD! % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWdXbqaamaalaaabaGaamyzamaaCaaaleqabaWaaOaaaeaacaWG % 4baameqaaaaaaOqaamaakaaabaGaamiEaaWcbeaaaaGccaWGKbGaam % iEaaWcbaGaaGymaaqaaiaaisdaa0Gaey4kIipaaaa!3F47! \int\limits_1^4 {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}dx} = \int\limits_1^2 {{e^u}.2du = } 2\int\limits_1^2 {{e^u}du} \)
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