Основные интегралы — различия между версиями
Материал из Вики ИТ мехмата ЮФУ
Juliet (обсуждение | вклад) |
Ulysses (обсуждение | вклад) (table) |
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− | <math>~\int\!0\, dx = C</math> | + | {| {{Integrals}} |
− | + | |- | |
− | <math>~\int\!\,dx = x +C</math> | + | |<math>~\int\!0\, dx = C</math> |
− | + | |<math>~\int\!\,dx = x +C</math> | |
− | + | |- | |
− | <math>~\int\!\sqrt{x}\,dx = {2 \over 3} x\sqrt{x} +C</math> | + | |<math>~\int\!\sqrt{x}\,dx = {2 \over 3} x\sqrt{x} +C</math> |
− | + | |<math>~\int\!x^n\,dx = \frac{x^{n+1}}{n+1} + C,\,n \ne -1</math> | |
− | <math>~\int\!x^n\,dx = \frac{x^{n+1}}{n+1} + C,\,n \ne -1</math> | + | |- |
− | + | |<math>\int {dx\over x}\, = \ln \left|x \right| + C</math> | |
− | + | |<math>\int {dx\over \sqrt{x}}\, = 2\sqrt{x} + C</math> | |
− | <math>\int {dx\over x}\, = \ln \left|x \right| + C</math> | + | |- |
− | + | |<math>\int {dx\over x^2}\, = {-1 \over x} + C</math> | |
− | <math>\int {dx\over \sqrt{x}}\, = 2\sqrt{x} + C</math> | + | |<math>\int\!e^x\,dx = e^x + C</math> |
− | + | |- | |
− | <math>\int {dx\over x^2}\, = {-1 \over x} + C</math> | + | |<math>\int\!a^x\,dx = \frac{a^x}{\ln{a}} + C</math> |
− | + | |<math>\int\!\sin{x}\, dx = -\cos{x} + C</math> | |
− | + | |- | |
− | <math>\int\!e^x\,dx = e^x + C</math> | + | |<math>\int\!\cos{x}\, dx = \sin{x} + C</math> |
− | + | |<math>\int\!{dx \over \cos^2 x} = \operatorname{tg}\,x + C</math> | |
− | <math>\int\!a^x\,dx = \frac{a^x}{\ln{a}} + C</math> | + | |- |
− | + | |<math>\int\!{dx \over \sin^2 x} = - \operatorname{ctg}\,x + C</math> | |
− | + | |<math>\int\!{dx \over {1+x^2}} = \operatorname{arctg}\,x + C</math> | |
− | <math>\int\!\sin{x}\, dx = -\cos{x} + C</math> | + | |- |
− | + | |<math>\int\!{dx \over \sqrt{1-x^2}} = \arcsin {x} + C</math> | |
− | <math>\int\!\cos{x}\, dx = \sin{x} + C</math> | + | |<math>\int\!{-dx \over \sqrt{1-x^2}} = \arccos {x} + C</math> |
− | + | |- | |
− | <math>\int\!{dx \over \cos^2 x} = \operatorname{tg}\,x + C</math> | + | |<math>\int \operatorname{sh}\,x \, dx = \operatorname{ch}\,x + C</math> |
− | + | |<math>\int \operatorname{ch}\,x \, dx = \operatorname{sh}\,x + C</math> | |
− | <math>\int\!{dx \over \sin^2 x} = - \operatorname{ctg}\,x + C</math> | + | |- |
− | + | |<math>\int\!{dx \over \operatorname{ch}^2 x} = \operatorname{th}\,x + C</math> | |
− | + | |<math>\int\!{dx \over \operatorname{sh}^2 x} = - \operatorname{cth}\,x + C</math> | |
− | <math>\int\!{dx \over {1+x^2}} = \operatorname{arctg}\,x + C</math> | + | |} |
− | |||
− | <math>\int\!{dx \over \sqrt{1-x^2}} = \arcsin {x} + C</math> | ||
− | |||
− | <math>\int\!{-dx \over \sqrt{1-x^2}} = \arccos {x} + C</math> | ||
− | |||
− | |||
− | <math>\int \operatorname{sh}\,x \, dx = \operatorname{ch}\,x + C</math> | ||
− | |||
− | <math>\int \operatorname{ch}\,x \, dx = \operatorname{sh}\,x + C</math> | ||
− | |||
− | <math>\int\!{dx \over \operatorname{ch}^2 x} = \operatorname{th}\,x + C</math> | ||
− | |||
− | <math>\int\!{dx \over \operatorname{sh}^2 x} = - \operatorname{cth}\,x + C</math> |
Версия 19:57, 22 марта 2009
<math>~\int\!0\, dx = C</math> | <math>~\int\!\,dx = x +C</math> |
<math>~\int\!\sqrt{x}\,dx = {2 \over 3} x\sqrt{x} +C</math> | <math>~\int\!x^n\,dx = \frac{x^{n+1}}{n+1} + C,\,n \ne -1</math> |
x \right| + C</math> | <math>\int {dx\over \sqrt{x}}\, = 2\sqrt{x} + C</math> |
<math>\int {dx\over x^2}\, = {-1 \over x} + C</math> | <math>\int\!e^x\,dx = e^x + C</math> |
<math>\int\!a^x\,dx = \frac{a^x}{\ln{a}} + C</math> | <math>\int\!\sin{x}\, dx = -\cos{x} + C</math> |
<math>\int\!\cos{x}\, dx = \sin{x} + C</math> | <math>\int\!{dx \over \cos^2 x} = \operatorname{tg}\,x + C</math> |
<math>\int\!{dx \over \sin^2 x} = - \operatorname{ctg}\,x + C</math> | <math>\int\!{dx \over {1+x^2}} = \operatorname{arctg}\,x + C</math> |
<math>\int\!{dx \over \sqrt{1-x^2}} = \arcsin {x} + C</math> | <math>\int\!{-dx \over \sqrt{1-x^2}} = \arccos {x} + C</math> |
<math>\int \operatorname{sh}\,x \, dx = \operatorname{ch}\,x + C</math> | <math>\int \operatorname{ch}\,x \, dx = \operatorname{sh}\,x + C</math> |
<math>\int\!{dx \over \operatorname{ch}^2 x} = \operatorname{th}\,x + C</math> | <math>\int\!{dx \over \operatorname{sh}^2 x} = - \operatorname{cth}\,x + C</math> |