\(\int \sin ^{3} x d x\)  bằng

Tính \(\int {\frac{{2x}}{{({x^2} + 9})^4}dx} \)

Biết \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % WcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWG4bGa % ey4kaSIaaGymaaqaaiaadIhacqGHRaWkcaaIXaaaaiaabsgacaWG4b % Gaeyypa0JaamyyaiabgUcaRiGacYgacaGGUbWaaSaaaeaacaWGIbaa % baGaaGOmaaaaaSqaaiaaiodaaeaacaaI1aaaniabgUIiYdaaaa!4A38! \int\limits_3^5 {\frac{{{x^2} + x + 1}}{{x + 1}}{\rm{d}}x = a + \ln \frac{b}{2}} \) với a, b là các số nguyên. Tính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9iaadkgadaahaaWcbeqaaiaaikdaaaGccqGHsislcaWGHbaaaa!3B7E! S = {b^2} - a\)

Tìm nguyên hàm  F(x) của hàm số .

Cho \( \smallint \sqrt[6]{{1 - {{\cos }^3}x}}\sin x{\cos ^5}xdx = 2\left( {\frac{{{t^\alpha }}}{\alpha } - \frac{{{t^\beta }}}{\beta }} \right)+C\) với \(t = \sqrt[6]{{1 - {{\cos }^3}x}}\)  . Tỉ số \( \frac{\alpha }{\beta }\)

Tìm nguyên hàm của hàm số f(x) = 2x +1.

Hàm số \(f\left( x \right) = {e^x}\left( {\ln 2 + \frac{{{e^{ - x}}}}{{{{\sin }^2}x}}} \right)\) có họ nguyên hàm là

Để tính \(\;\smallint x\ln \left( {2\; + \;x} \right).dx\) theo phương pháp tính nguyên hàm từng phần, ta đặt:

Tìm họ nguyên hàm F(x) của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpciGG % ZbGaaiyAaiaac6gacaaIYaGaamiEaiabgUcaRiaaikdacaWG4baaaa!439A! y = f\left( x \right) = \sin 2x + 2x\)

Họ nguyên hàm của hàm số \(f(x)=\mathrm{e}^{x}+x \) là?

Tìm nguyên hàm: \(K = \smallint \frac{{\ln x\sqrt[3]{{2 + {{\ln }^2}x}}}}{x}dx.\)

\(\text { Cho hàm số } F(x)=\int x \sqrt{x^{2}+2} \mathrm{~d} x \text { . Biết } F(\sqrt{2})=\frac{2}{3}, \text { tính } F(\sqrt{7}) \text { . }\)

Cho \(\int f(4 x) \mathrm{d} x=e^{2 x}-x^{2}+C\). Khi đó \(\int f(-x) \mathrm{d} x\) bằng:

Tìm nguyên hàm của hàm số \(f\left( x \right) = \sqrt[3]{{x - 2}}\)

Cho F(x) là một nguyên hàm của hàm số \(f(x)=e^{x}+2 x\) thỏa mãn \(F(0)=\frac{3}{2}\) . Khi đó F(x) là:

Tìm nguyên hàm của hàm số \(f(x)=e^{x}-e^{-x}\)

Tìm nguyên hàm của hàm số \(f(x)=\sqrt[3]{x-2}\)

Tìm nguyên hàm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9maapeaabaGaamiEaiGacogacaGGVbGaai4CaiaadIhacaqGKbGa % amiEaaWcbeqab0Gaey4kIipaaaa!4074! I = \int {x\cos x{\rm{d}}x} \)

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 

Tìm tất cả các tham số thực m > 1 để phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqaMzfvLHfij5gC1rhimfMBNvxyNvga7XvAUrhxSL % wBPr3CFThm951ET13ECXwzMrhkGidETediCjxANHgDPWfDLHhD7rwF % 41xp7ThE951EY0xFTmdERqtFamXvP5wqSXMqHnxAJn0BKvguHDwzZb % qefqvATv2CG4uz3bIuV1wyUbqedmvETj2BSbqefm0B1jxALjhiov2D % aebbnrfifHhDYfgasaacH8YjY-vipgYlh9vqqj-hEeeu0xXdbba9fr % Fj0-OqFfea0dXdd9vqai-hGuQ8kuc9pgc9q8qqaq-dir-f0-yqaiVg % Fr0xfr-xfr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaa % aaaaaaa8qadaWdXbWdaeaapeWaaeWaa8aabaWdbiaaikdacaWG4bGa % eyOeI0IaaGymaaGaayjkaiaawMcaaiaabsgacaWG4baal8aabaWdbi % aaicdaa8aabaWdbiaad2gaa0Gaey4kIipakiabg2da9iaadIhapaWa % aWbaaSqabeaapeGaaGOmaaaakiabgkHiTiaaiodacaWG4bGaey4kaS % IaaGinaaaa!6951! \int\limits_0^m {\left( {2x - 1} \right){\rm{d}}x} = {x^2} - 3x + 4\) có hai nghiệm phân biệt?

Nguyên hàm của hàm số \(y = sin^2x.cos^3x\)là