Tìm một nguyên hàm F(x) của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadggacaWG4bGaey4k % aSYaaSaaaeaacaWGIbaabaGaamiEamaaCaaaleqabaGaaGOmaaaaaa % aaaa!400D! f\left( x \right) = ax + \frac{b}{{{x^2}}}\) \(x \ne 0\) biết rằng F(1) = 4; F(-1) = 1; f(1) =0

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaG4maiaa % dIhadaahaaWcbeqaaiaaikdaaaaakeaacaaI0aaaaiabgUcaRmaala % aabaGaaG4maaqaaiaaikdacaWG4baaaiabgUcaRmaalaaabaGaaG4n % aaqaaiaaisdaaaaaaa!439F! F\left( x \right) = \frac{{3{x^2}}}{4} + \frac{3}{{2x}} + \frac{7}{4}\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaG4maiaa % dIhadaahaaWcbeqaaiaaikdaaaaakeaacaaI0aaaaiabgkHiTmaala % aabaGaaG4maaqaaiaaikdacaWG4baaaiabgkHiTmaalaaabaGaaG4n % aaqaaiaaisdaaaaaaa!43B5! F\left( x \right) = \frac{{3{x^2}}}{4} - \frac{3}{{2x}} - \frac{7}{4}\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaG4maiaa % dIhadaahaaWcbeqaaiaaikdaaaaakeaacaaIYaaaaiabgUcaRmaala % aabaGaaG4maaqaaiaaisdacaWG4baaaiabgkHiTmaalaaabaGaaG4n % aaqaaiaaisdaaaaaaa!43AA! F\left( x \right) = \frac{{3{x^2}}}{2} + \frac{3}{{4x}} - \frac{7}{4}\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaG4maiaa % dIhadaahaaWcbeqaaiaaikdaaaaakeaacaaIYaaaaiabgkHiTmaala % aabaGaaG4maaqaaiaaikdacaWG4baaaiabgkHiTmaalaaabaGaaGym % aaqaaiaaikdaaaaaaa!43AB! F\left( x \right) = \frac{{3{x^2}}}{2} - \frac{3}{{2x}} - \frac{1}{2}\)

Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án

Môn: Toán Lớp 12

Chủ đề: Nguyên Hàm - Tích Phân Và Ứng Dụng

Bài: Nguyên hàm

Lời giải:

Báo sai

Ta có họ các nguyên hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadggacaWG4bGaey4k % aSYaaSaaaeaacaWGIbaabaGaamiEamaaCaaaleqabaGaaGOmaaaaaa % aaaa!400D! f\left( x \right) = ax + \frac{b}{{{x^2}}}; x\ne 0\) có dạng: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaamyyaiaa % dIhadaahaaWcbeqaaiaaikdaaaaakeaacaaIYaaaaiabgkHiTmaala % aabaGaamOyaaqaaiaadIhaaaGaey4kaSIaam4qaaaa!4278! F\left( x \right) = \frac{{a{x^2}}}{2} - \frac{b}{x} + C\)

Theo giả thiết ta có hệ: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaamaalaaabaGaamyyaaqaaiaaikdaaaGaey4kaSIaamOyaiabgUca % RiaadoeacqGH9aqpcaaIXaaabaWaaSaaaeaacaWGHbaabaGaaGOmaa % aacqGHsislcaWGIbGaey4kaSIaam4qaiabg2da9iaaisdaaeaacaWG % HbGaey4kaSIaamOyaiabg2da9iaaicdaaaGaay5Eaaaaaa!495F! \left\{ \begin{array}{l} \frac{a}{2} + b + C = 1\\ \frac{a}{2} - b + C = 4\\ a + b = 0 \end{array} \right.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaadggacqGH9aqpdaWcaaqaaiaaiodaaeaacaaIYaaa % aaqaaiaadkgacqGH9aqpcqGHsisldaWcaaqaaiaaiodaaeaacaaIYa % aaaaqaaiaadoeacqGH9aqpdaWcaaqaaiaaiEdaaeaacaaI0aaaaaaa % caGL7baaaaa!44A7! \Leftrightarrow \left\{ \begin{array}{l} a = \frac{3}{2}\\ b = - \frac{3}{2}\\ C = \frac{7}{4} \end{array} \right.\)

. Từ đó hàm số f(x) có một nguyên hàm là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaG4maiaa % dIhadaahaaWcbeqaaiaaikdaaaaakeaacaaI0aaaaiabgUcaRmaala % aabaGaaG4maaqaaiaaikdacaWG4baaaiabgUcaRmaalaaabaGaaG4n % aaqaaiaaisdaaaaaaa!439F! F\left( x \right) = \frac{{3{x^2}}}{4} + \frac{3}{{2x}} + \frac{7}{4}\).