Tavola degli integrali indefiniti di funzioni razionali

Da Wikipedia, l'enciclopedia libera.

Questa pagina contiene una tavola di integrali indefiniti di funzioni razionali.

Per altri integrali vedi Indici per la matematica#Tavole di integrali.

$\int (ax + b)^n dx = \frac{(ax + b)^{n+1}}{a(n + 1)} \qquad\mbox{(per } n\neq -1\mbox{)}$

$\int\frac{dx}{ax + b} = \frac{1}{a}\ln\left|ax + b\right|$

$\int x(ax + b)^n dx = \frac{a(n + 1)x - b}{a^2(n + 1)(n + 2)} (ax + b)^{n+1} \qquad\mbox{(per }n \not\in \{-1, -2\}\mbox{)}$

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

$\int\frac{x\;dx}{(ax + b)^2} = \frac{b}{a^2(ax + b)} + \frac{1}{a^2}\ln\left|ax + b\right|$

$\int\frac{x\;dx}{(ax + b)^n} = \frac{a(1 - n)x - b}{a^2(n - 1)(n - 2)(ax + b)^{n-1}} \qquad\mbox{(per } n\not\in \{-1, -2\}\mbox{)}$

$\int\frac{x^2\;dx}{ax + b} = \frac{1}{a^3}\left(\frac{(ax + b)^2}{2} - 2b(ax + b) + b^2\ln\left|ax + b\right|\right)$

$\int\frac{x^2\;dx}{(ax + b)^2} = \frac{1}{a^3}\left(ax + b - 2b\ln\left|ax + b\right| - \frac{b^2}{ax + b}\right)$

$\int\frac{x^2\;dx}{(ax + b)^3} = \frac{1}{a^3}\left(\ln\left|ax + b\right| + \frac{2b}{ax + b} - \frac{b^2}{2(ax + b)^2}\right)$

$\int\frac{x^2\;dx}{(ax + b)^n} = \frac{1}{a^3}\left(-\frac{1}{(n- 3)(ax + b)^{n-3}} + \frac{2b}{(n-2)(a + b)^{n-2}} - \frac{b^2}{(n - 1)(ax + b)^{n-1}}\right) \qquad\mbox{(per } n\not\in \{1, 2, 3\}\mbox{)}$

$\int\frac{dx}{x(ax + b)} = -\frac{1}{b}\ln\left|\frac{ax+b}{x}\right|$

$\int\frac{dx}{x^2(ax+b)} = -\frac{1}{bx} + \frac{a}{b^2}\ln\left|\frac{ax+b}{x}\right|$

$\int\frac{dx}{x^2(ax+b)^2} = -a\left(\frac{1}{b^2(ax+b)} + \frac{1}{ab^2x} - \frac{2}{b^3}\ln\left|\frac{ax+b}{x}\right|\right)$

$\int\frac{dx}{x^2+a^2} = \frac{1}{a}\arctan\frac{x}{a}$

$\int\frac{dx}{x^2-a^2} = -\frac{1}{a}\,\mathrm{artanh}\frac{x}{a} = \frac{1}{2a}\ln\frac{a-x}{a+x} \qquad\mbox{(per }|x| < |a|\mbox{)}$

$\int\frac{dx}{x^2-a^2} = -\frac{1}{a}\,\mathrm{arcoth}\frac{x}{a} = \frac{1}{2a}\ln\frac{x-a}{x+a} \qquad\mbox{(per }|x| > |a|\mbox{)}$

$\int\frac{dx}{ax^2+bx+c} = \frac{2}{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} \qquad\mbox{(per }4ac-b^2>0\mbox{)}$

$\int\frac{dx}{ax^2+bx+c} = -\frac{2}{\sqrt{b^2-4ac}}\,\mathrm{arctanh}\frac{2ax+b}{\sqrt{b^2-4ac}} = \frac{1}{\sqrt{b^2-4ac}}\ln\left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right| \qquad\mbox{(per }4ac-b^2<0\mbox{)}$

$\int\frac{x\;dx}{ax^2+bx+c} = \frac{1}{2a}\ln\left|ax^2+bx+c\right|-\frac{b}{2a}\int\frac{dx}{ax^2+bx+c}$

$\int\frac{mx+n}{ax^2+bx+c}dx = \frac{m}{2a}\ln\left|ax^2+bx+c\right|+\frac{2an-bm}{a\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} \qquad\mbox{(per }4ac-b^2>0\mbox{)}$

$\int\frac{mx+n}{ax^2+bx+c}dx = \frac{m}{2a}\ln\left|ax^2+bx+c\right|+\frac{2an-bm}{a\sqrt{b^2-4ac}}\,\mathrm{artanh}\frac{2ax+b}{\sqrt{b^2-4ac}} \qquad\mbox{(per }4ac-b^2<0\mbox{)}$