Предмет: Алгебра,
автор: Аноним
ПОМОГИТЕ ПОЖАЛУЙСТА 6 ИНТЕГРАЛОВ 100 БАЛЛОВ!!!!!!!!!!!!!!!
Приложения:

Ответы
Автор ответа:
1
(1)

(2)

(3)
![\int\limits {\frac{x}{\sqrt{3-2x^2}}} \, dx = \int\limits {\frac{2x}{2\sqrt{3-2x^2}}} \, dx = \int\limits {\frac{1}{2\sqrt{3-2x^2}}} \, d(x^2) =\\\\=
\int\limits {\frac{1}{-4\sqrt{3-2x^2}}} \, d(-2x^2)=\int\limits {\frac{1}{-4\sqrt{3-2x^2}}} \, d(3-2x^2)=\\\\
=[t=3-2x^2]=-\frac{1}{4}* \int\limits {\frac{1}{\sqrt{t}}} \, dt=-\frac{1}{4}* \int\limits {t^{-\frac{1}{2}}} \, dt=\\\\
=-\frac{1}{4}*\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C=-\frac{\sqrt{t}}{2}+C=-\frac{\sqrt{3-2x^2}}{2}+C \int\limits {\frac{x}{\sqrt{3-2x^2}}} \, dx = \int\limits {\frac{2x}{2\sqrt{3-2x^2}}} \, dx = \int\limits {\frac{1}{2\sqrt{3-2x^2}}} \, d(x^2) =\\\\=
\int\limits {\frac{1}{-4\sqrt{3-2x^2}}} \, d(-2x^2)=\int\limits {\frac{1}{-4\sqrt{3-2x^2}}} \, d(3-2x^2)=\\\\
=[t=3-2x^2]=-\frac{1}{4}* \int\limits {\frac{1}{\sqrt{t}}} \, dt=-\frac{1}{4}* \int\limits {t^{-\frac{1}{2}}} \, dt=\\\\
=-\frac{1}{4}*\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C=-\frac{\sqrt{t}}{2}+C=-\frac{\sqrt{3-2x^2}}{2}+C](https://tex.z-dn.net/?f=+%5Cint%5Climits+%7B%5Cfrac%7Bx%7D%7B%5Csqrt%7B3-2x%5E2%7D%7D%7D+%5C%2C+dx+%3D+%5Cint%5Climits+%7B%5Cfrac%7B2x%7D%7B2%5Csqrt%7B3-2x%5E2%7D%7D%7D+%5C%2C+dx+%3D+%5Cint%5Climits+%7B%5Cfrac%7B1%7D%7B2%5Csqrt%7B3-2x%5E2%7D%7D%7D+%5C%2C+d%28x%5E2%29+%3D%5C%5C%5C%5C%3D%0A%5Cint%5Climits+%7B%5Cfrac%7B1%7D%7B-4%5Csqrt%7B3-2x%5E2%7D%7D%7D+%5C%2C+d%28-2x%5E2%29%3D%5Cint%5Climits+%7B%5Cfrac%7B1%7D%7B-4%5Csqrt%7B3-2x%5E2%7D%7D%7D+%5C%2C+d%283-2x%5E2%29%3D%5C%5C%5C%5C%0A%3D%5Bt%3D3-2x%5E2%5D%3D-%5Cfrac%7B1%7D%7B4%7D%2A+%5Cint%5Climits+%7B%5Cfrac%7B1%7D%7B%5Csqrt%7Bt%7D%7D%7D+%5C%2C+dt%3D-%5Cfrac%7B1%7D%7B4%7D%2A+%5Cint%5Climits+%7Bt%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%7D+%5C%2C+dt%3D%5C%5C%5C%5C%0A%3D-%5Cfrac%7B1%7D%7B4%7D%2A%5Cfrac%7Bt%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B%5Cfrac%7B1%7D%7B2%7D%7D%2BC%3D-%5Cfrac%7B%5Csqrt%7Bt%7D%7D%7B2%7D%2BC%3D-%5Cfrac%7B%5Csqrt%7B3-2x%5E2%7D%7D%7B2%7D%2BC)
(4)

(5)

(6)
![\int\limits {e^{-3x}} \, dx = \int\limits {\frac{e^{-3x}}{-3}} \, d(-3x)
=[t=-3x]=\\\\=-\frac{1}{3}* \int\limits {e^t} \, dt=-\frac{1}{3}e^t+C=\\\\
=-\frac{1}{3}e^{-3x}+C \int\limits {e^{-3x}} \, dx = \int\limits {\frac{e^{-3x}}{-3}} \, d(-3x)
=[t=-3x]=\\\\=-\frac{1}{3}* \int\limits {e^t} \, dt=-\frac{1}{3}e^t+C=\\\\
=-\frac{1}{3}e^{-3x}+C](https://tex.z-dn.net/?f=+%5Cint%5Climits+%7Be%5E%7B-3x%7D%7D+%5C%2C+dx+%3D+%5Cint%5Climits+%7B%5Cfrac%7Be%5E%7B-3x%7D%7D%7B-3%7D%7D+%5C%2C+d%28-3x%29+%0A%3D%5Bt%3D-3x%5D%3D%5C%5C%5C%5C%3D-%5Cfrac%7B1%7D%7B3%7D%2A+%5Cint%5Climits+%7Be%5Et%7D+%5C%2C+dt%3D-%5Cfrac%7B1%7D%7B3%7De%5Et%2BC%3D%5C%5C%5C%5C%0A%3D-%5Cfrac%7B1%7D%7B3%7De%5E%7B-3x%7D%2BC)
(2)
(3)
(4)
(5)
(6)
Интересные вопросы
Предмет: Физика,
автор: sosksbk
Предмет: Математика,
автор: duusshamooya
Предмет: Обществознание,
автор: aahahahahaa
Предмет: Математика,
автор: arina8694
Предмет: Математика,
автор: MaxSokolovich