Uniwersalne podstawienia:

$${t = \operatorname{tg}\frac{x}{2}}{x = 2*arctg\left( t \right)}{dx = \frac{2}{t^{2} + 1}dt}{sinx = \frac{2t}{t^{2} + 1}}{cosx = \frac{1 - t^{2}}{1 + t^{2}}}$$

$${t = tg\ x}{x = arctg\ t\backslash n}{dx = \frac{\text{dt}}{1 + t^{2}}\backslash n}{\sin^{2}x = \frac{t^{2}}{t^{2} + 1}\backslash n}{\cos^{2}x = \frac{1}{t^{2} + 1}\backslash n}{sinxcosx = \frac{t}{t^{2} + 1}}$$

Podstawienia Eulera: $R\left( \sqrt{ax^{2} + bx + c},\ x \right)$

$${\sqrt{ax^{2} + bx + c} = t - \sqrt{a}\text{\ x}\backslash n}{x = \frac{t^{2} - c}{b + 2\sqrt{a}\text{\ t}}\backslash n}{dx = \frac{2t\left( b + 2\sqrt{a}\text{\ t} \right) - 2\sqrt{a}\left( t^{2} - c \right)}{\left( b + 2\sqrt{a}\text{\ t} \right)^{2}}\text{dt}\backslash n}{\sqrt{ax^{2} + bx + c} = \sqrt{a}\frac{t^{2} - c}{b - 2\sqrt{a}t} + t}$$

$${\sqrt{ax^{2} + bx + c} = xt + \sqrt{c}\backslash n}{x = \frac{2\sqrt{c}t - b}{a - t^{2}}\backslash n}{dx = \frac{2\sqrt{c}\left( a - t^{2} \right) + 2t\left( 2\sqrt{c}t - b \right)}{\left( a - t^{2} \right)^{2}}\text{dt}\backslash n}{\sqrt{ax^{2} + bx + c} = \frac{2\sqrt{c}t^{2} - bt}{a - t^{2}} + \sqrt{c}}$$

$${\sqrt{ax^{2} + bx + c} = \sqrt{a\left( x - x_{0} \right)\left( x - x_{1} \right)} = t\left( x - x_{1} \right)\backslash n}{\left( x - x_{1} \right)t^{2} = a\left( x - x_{0} \right) = > x = \frac{t^{2}x_{1} - ax_{0}}{t^{2} - a}\backslash n}{dx = \frac{2ta\left( x_{0} - x_{1} \right)}{\left( t^{2} - a \right)^{2}}\text{dt}\backslash n}{\sqrt{ax^{2} + bx + c} = t\left( \frac{t^{2}x_{1} - ax_{0}}{t^{2} - a} - x_{1} \right)}$$

$${\int_{}^{}{\frac{1}{\sqrt{x^{2} + 1}}dx = \ln\left( x + \sqrt{x^{2} + 1} \right) + C}\backslash n}{\int_{}^{}{\frac{1}{\sqrt{{1 - x}^{2}}}dx = \arcsin\left( x \right) + C}\backslash n}{\int_{}^{}{\frac{1}{\sqrt{x^{2} - 1}} = \ln\left| x + \sqrt{x^{2} - 1} \right| + C}\backslash n}$$