Списак интеграла ирационалних функција

Списак интеграла ирационалних функција:

Интеграли са -{a² - x²}-

$\int \sqrt{a^2-x^2}dx=\frac{1}{2}(x\sqrt{a^2-x^2}+a^2\arcsin \frac{x}{a})\text{(}|x|\le |a|\text{)}$

$\int x\sqrt{a^2-x^2}dx=-\frac{1}{3}\sqrt{(a^2-x^2{)}^3}\text{(}|x|\le |a|\text{)}$

$\int \frac{\sqrt{a^2-x^2}dx}{x}=\sqrt{a^2-x^2}-a\ln \left|\frac{a+\sqrt{a^2+x^2}}{x}\right|\text{(}|x|\le |a|\text{)}$

$\int \frac{dx}{\sqrt{a^2-x^2}}=\arcsin \frac{x}{a}\text{(}|x|\le |a|\text{)}$

$\int \frac{x^{2}dx}{\sqrt{a^2-x^2}}=-\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\arcsin \frac{x}{a}\text{(}|x|\le |a|\text{)}$

Интеграли са -{x² + a²}-

$\int \sqrt{x^2+a^2}dx=\frac{1}{2}(x\sqrt{x^2+a^2}+a^2\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}\frac{x}{a})$

$\int x\sqrt{x^2+a^2}dx=\frac{1}{3}\sqrt{(x^2+a^2{)}^3}$

$\int \frac{\sqrt{x^2+a^2}dx}{x}=\sqrt{x^2+a^2}-a\ln \left|\frac{a+\sqrt{x^2+a^2}}{x}\right|$

$\int \frac{dx}{\sqrt{x^2+a^2}}=\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}\frac{x}{a}=\ln |x+\sqrt{x^2+a^2}|$

$\int \frac{xdx}{\sqrt{x^2+a^2}}=\sqrt{x^2+a^2}$

$\int \frac{x^{2}dx}{\sqrt{x^2+a^2}}=\frac{x}{2}\sqrt{x^2+a^2}-\frac{a^2}{2}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}\frac{x}{a}=\frac{x}{2}\sqrt{x^2+a^2}-\frac{a^2}{2}\ln |x+\sqrt{x^2+a^2}|$

$\int \frac{dx}{x\sqrt{x^2+a^2}}=-\frac{1}{a}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}\frac{a}{x}=-\frac{1}{a}\ln \left|\frac{a+\sqrt{x^2+a^2}}{x}\right|$

Интеграли са -{x² - a²}-

$\int \sqrt{x^2-a^2}dx=\frac{1}{2}(x\sqrt{x^2-a^2}-\mathrm{sgn}x\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}\left|\frac{x}{a}\right|)\text{(za }|x|\ge |a|\text{)}$

$\int x\sqrt{x^2-a^2}dx=\frac{1}{3}\sqrt{(x^2-a^2{)}^3}\text{(za }|x|\ge |a|\text{)}$

$\int \frac{\sqrt{x^2-a^2}dx}{x}=\sqrt{x^2-a^2}-a\arccos \frac{a}{x}\text{(za }|x|\ge |a|\text{)}$

$\int \frac{dx}{\sqrt{x^2-a^2}}=\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}\frac{x}{a}=\ln (|x|+\sqrt{x^2-a^2})\text{(za }|x|>|a|\text{)}$

$\int \frac{xdx}{\sqrt{x^2-a^2}}=\sqrt{x^2-a^2}\text{(za }|x|>|a|\text{)}$

$\int \frac{x^{2}dx}{\sqrt{x^2-a^2}}=\frac{x}{2}\sqrt{x^2-a^2}+\frac{a^2}{2}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}\left|\frac{x}{a}\right|=\frac{1}{2}(x\sqrt{x^2-a^2}+a^2\ln (|x|+\sqrt{x^2-a^2}))\text{(za }|x|>|a|\text{)}$

Интеграли са -{ax² + bx + c}-

$\int \sqrt{a{x}^2+bx+c}dx=\frac{1}{8a^{3/2}}(2\sqrt{a}\cdot (b+2ax)\sqrt{c+x(b+ax)}-(b^2-4ac)\log (b+2x+2\sqrt{a}\sqrt{c+x(b+ax)}))$

$\int \frac{dx}{\sqrt{a{x}^2+bx+c}}=\frac{1}{\sqrt{a}}\ln |2\sqrt{a(a{x}^2+bx+c)}+2ax+b|\text{(za }a>0\text{)}$

$\int \frac{dx}{\sqrt{a{x}^2+bx+c}}=\frac{1}{\sqrt{a}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}\frac{2ax+b}{\sqrt{4ac-b^2}}\text{(za }a>0\text{, }4ac-b^2>0\text{)}$

$\int \frac{dx}{\sqrt{a{x}^2+bx+c}}=\frac{1}{\sqrt{a}}\ln |2ax+b|\text{(za }a>0\text{, }4ac-b^2=0\text{)}$

$\int \frac{dx}{\sqrt{a{x}^2+bx+c}}=-\frac{1}{\sqrt{-a}}\arcsin \frac{2ax+b}{\sqrt{b^2-4ac}}\text{(za }a<0\text{, }4ac-b^2<0\text{)}$

$\int \frac{xdx}{\sqrt{a{x}^2+bx+c}}=\frac{\sqrt{a{x}^2+bx+c}}{a}-\frac{b}{2a}\int \frac{dx}{\sqrt{a{x}^2+bx+c}}$

Литература

Шаблон:Литература-{

• Шаблон:Cite book
• Шаблон:Cite book. Errata. (Several previous editions as well.)
• A.P. Prudnikov (А. П. Прудников), Yu.A. Brychkov (Ю. А. Брычков), O.I. Marichev (О. И. Маричев). Integrals and Series. First edition (Russian), volume 1–5, Nauka, 1981−1986. First edition (English, translated from the Russian by N.M. Queen), volume 1–5, Gordon & Breach Science Publishers/CRC Press, 1988–Шаблон:Page. Second revised edition (Russian), volume 1–3, Fiziko-Matematicheskaya Literatura, 2003.
• Yu.A. Brychkov (Ю. А. Брычков). Handbook of Special Functions: Derivatives, Integrals, Series and Other Formulas. Russian edition, Fiziko-Matematicheskaya Literatura, 2006. English edition. Шаблон:Page1.
• Шаблон:Cite book. (Many earlier editions as well.)
• Meyer Hirsch, Integral Tables, Or, A Collection of Integral Formulae (Baynes and son, London, 1823) [English translation of Integraltafeln]
• Benjamin O. Pierce A short table of integrals - revised edition (Ginn & co., Boston, 1899)

}-Шаблон:Литература крај