Tìm một nguyên hàm F(x) của hàm số \(f(x)=\sin ^{2} \frac{x}{2} \text { biết } F\left(\frac{\pi}{2}\right)=\frac{\pi}{4}\)

Cho F(x) là một nguyên hàm của \(f(x)=\frac{1}{x-1} \text { trên khoảng }(1 ;+\infty) \text { thỏa mãn } F(\mathrm{e}+1)=4\). Tìm F(x)

Nếu có x = cot t thì:

Họ nguyên hàm của hàm số \(f(x)=\sin 5 x+2\) là?

Nguyên hàm \(\int[\sin (2 x+3)+\cos (3-2 x)]dx\)

F(x) là một nguyên hàm của hàm số \(f\left( x \right) = \frac{{2x + 3}}{{{x^2}}}\), biết rằng F(1) = 1. F(x) là biểu thức nào sau đây

Cho biết \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaGymaaqa % aiaaiodaaaGaamiEamaaCaaaleqabaGaaG4maaaakiabgUcaRiaaik % dacaWG4bGaeyOeI0YaaSaaaeaacaaIXaaabaGaamiEaaaaaaa!4314! F\left( x \right) = \frac{1}{3}{x^3} + 2x - \frac{1}{x}\) là một nguyên hàm của \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaWaaeWaaeaa % caWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamyyaaGaayjkai % aawMcaamaaCaaaleqabaGaaGOmaaaaaOqaaiaadIhadaahaaWcbeqa % aiaaikdaaaaaaaaa!4295! f\left( x \right) = \frac{{{{\left( {{x^2} + a} \right)}^2}}}{{{x^2}}}\). Tìm nguyên hàm của \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadIhaciGGJbGaai4B % aiaacohacaWGHbGaamiEaaaa!401F! g\left( x \right) = x\cos ax\)

\(\text { Tìm nguyên hàm } F(x) \text { của hàm số } f(x)=6 x+\sin 3 x \text { , biết } F(0)=\frac{2}{3} \text { . }\)

\(\text { Hàm số } f(x) \text { thỏa mãn } f^{\prime}(x)=x e^{x} \text { là }\)

Cho F(x) là một nguyên hàm của hàm số \(f(x)=x^{2}-2 x+3\) thỏa mãn F(0)=2, giá trị của F (1) bằng

Nguyên hàm F(x) của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqaMrdvLHfij5gC1rhimfMBNvxyNvgaMXfBLzgDOa % cEGWLCPDgA0LspCzMCHn2EX03EYG3kX0hatCvAUfeBSjuyZL2yd9gz % LbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYL % wzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9 % v8qqaqFr0xc9pk0xbba9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0F % b9pgeaYRXxe9vr0-vr0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaa % aOqaaabaaaaaaaaapeGaamOzamaabmaapaqaa8qacaWG4baacaGLOa % GaayzkaaGaeyypa0ZaaSaaa8aabaWdbiaaigdaa8aabaWdbiaaikda % caWG4bGaey4kaSIaaGymaaaaaaa!5169! f\left( x \right) = \frac{1}{{2x + 1}}\), biết \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqaMrevLHfij5gC1rhimfMBNvxyNvgagXfBLzgDOa % cxMjxyJThx0vgE0Txz91sm9TNm9bcxYL2zOrxk9WLzYf2y7ntF7jtF % amXvP5wqSXMqHnxAJn0BKvguHDwzZbqefqvATv2CG4uz3bIuV1wyUb % qedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8Yj % Y-vipgYlh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqai-hGu % Q8kuc9pgc9q8qqaq-dir-f0-yqaiVgFr0xfr-xfr-xb9adbaqaaeGa % ciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8qacaWGgbWaaeWaa8 % aabaWdbmaalaaapaqaa8qacaqGLbGaeyOeI0IaaGymaaWdaeaapeGa % aGOmaaaaaiaawIcacaGLPaaacqGH9aqpdaWcaaWdaeaapeGaaG4maa % WdaeaapeGaaGOmaaaaaaa!57EA! F\left( {\frac{{{\rm{e}} - 1}}{2}} \right) = \frac{3}{2}\) là:

Nếu \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaKaaah % aakmaakaaajaaObaGaamiEaaqcbaAabaqcaaQaaeiiaKaaalaabsga % caWG4baajqwaacqaaiaadggaaeaacaWGIbaajmaWcqGHRiI8aKaaal % abg2da9OWaaSaaaKaaahaacaaIYaaabaGaaG4maaaaaaa!4722! \int\limits_a^b {\sqrt x {\rm{ d}}x} = \frac{2}{3}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaqadaWdaeaapeGaamyyaiabgwMiZkaaicdacaGGSaGaaGPaVlaa % ykW7caWGIbGaeyyzImRaaGimaaGaayjkaiaawMcaaaaa!424F! \left( {a \ge 0,\,\,b \ge 0} \right)\) thì 

Giả sử \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGacaGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % WcaaqaaiaaigdaaeaacaaIYaGaamiEaiabgUcaRiaaigdaaaGaaeiz % aiaadIhaaSqaaiaaigdaaeaacaaIYaaaniabgUIiYdGccqGH9aqpci % GGSbGaaiOBamaakaaabaWaaSaaaeaacaWGHbaabaGaamOyaaaaaSqa % baaaaa!44C3! \int\limits_1^2 {\frac{1}{{2x + 1}}{\rm{d}}x} = \ln \sqrt {\frac{a}{b}} \) với \(a,b \in N^*\) và a,b < 10 . TÍnh \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGacaGaaiaabeqaamaabaabaaGcbaGaamytaiabg2 % da9iaadggacqGHRaWkcaWGIbWaaWbaaSqabeaacaaIYaaaaaaa!3B62! M = a + {b^2}\)

Nguyên hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGacaGaaiaabeqaamaabaabaaGcbaGaamOzaiaacI % cacaWG4bGaaiykaiabg2da9iaaiodacaWG4bWaaWbaaSqabeaacaaI % YaaaaOGaey4kaSIaaGinaiaadIhacqGHRaWkdaWcaaqaaiaaigdaae % aacaWG4baaaaaa!422D! f(x) = 3{x^2} + 4x + \frac{1}{x}\) là:

Họ nguyên hàm của hàm số \(f(x)=(2 x+1) \ln x\) là:

Hàm số nào sau đây không là nguyên hàm của hàm số \(f\left( x \right) = \frac{{{x^2} + 2x}}{{{{\left( {x + 1} \right)}^2}}}\)

Tìm họ nguyên hàm của hàm số \(f(x)=3^{x}+\frac{1}{x^{2}}\)

Họ nguyên hàm của hàm số \(f\left( x \right) = \frac{{{e^x}}}{{{e^x} + 3}}\) là:

Tính nguyên hàm \(\int {\left( {\frac{1}{{2x + 3}}} \right)dx} \)

Tính \(\int\left(2+e^{3 x}\right)^{2} d x\).

Giả sử \(\int e^{2 x}\left(2 x^{3}+5 x^{2}-2 x+4\right) d x=\left(a x^{3}+b x^{2}+c x+d\right) e^{2 x}+C\) . Khi đó a+b+c+d bằng