CAŁKI

∫(f(x)+ g(x))dx =  ∫f(x)dx +  ∫g(x)dx

∫a f(x)dx = a ∫f(x)dx

CAŁKOWANIE PRZEZ PODSTAWIENIE:

$$\int_{}^{}\frac{\mathbf{f\ '(x)}}{\mathbf{\text{f\ }}\left( \mathbf{x} \right)}\mathbf{dx = \ ln\ }\left| \mathbf{\text{\ f}}\left( \mathbf{x} \right) \right|\mathbf{+ \ C}$$

CAŁKOWANIE PRZEZ CZĘŚĆI:

∫udv = uv −  ∫vdu

INNE WZORY:

$$\int_{}^{}\frac{\mathbf{\text{dx}}}{\mathbf{a}^{\mathbf{2}}\mathbf{+ \ }\mathbf{x}^{\mathbf{2}}}\mathbf{= \ }\frac{\mathbf{1}}{\mathbf{a}}\mathbf{\text{\ arctg\ }}\frac{\mathbf{x}}{\mathbf{a}}\mathbf{+ \ C}$$

$$\int_{}^{}\frac{\mathbf{\text{dx}}}{\sqrt{\mathbf{a}^{\mathbf{2}}\mathbf{- \ }\mathbf{x}^{\mathbf{2}}}}\mathbf{= \ }\mathbf{\arcsin}\mathbf{\ }\frac{\mathbf{x}}{\mathbf{a}}\mathbf{+ \ }\mathbf{C}$$

$$\int_{}^{}{\mathbf{cos\ ax\ dx = \ }\frac{\mathbf{1}}{\mathbf{a}}}\mathbf{\ sin\ ax + C}$$

$$\int_{}^{}{\mathbf{sin\ ax\ dx = - \ }\frac{\mathbf{1}}{\mathbf{a}}}\mathbf{\ cos\ ax + C}$$

$$\int_{}^{}{\mathbf{\text{e\ }}^{\mathbf{\text{ax}}}\mathbf{\ dx = \ }\frac{\mathbf{1}}{\mathbf{a}}}\mathbf{\ }\mathbf{e}^{\mathbf{\text{ax}}}\mathbf{+ C}$$

$$\int_{}^{}{\mathbf{\text{f\ }}\left( \mathbf{ax + b} \right)\mathbf{dx = \ }\frac{\mathbf{1}}{\mathbf{a}}}\mathbf{\text{\ F\ }}\left( \mathbf{ax + b} \right)\mathbf{+ \ C}$$

całka	f-cja pierwotna	całka	f-cja pierwotna
∫0 dx	C	∫x^αdx	$\frac{x^{\alpha + 1}}{\alpha + 1}$ + C, α ∈ R\ {-1}
$$\int_{}^{}{\frac{1}{x}dx = \ \int_{}^{}{x^{- 1}\text{dx}}}$$	ln|x| + C	∫a^xdx	$\frac{a^{x}}{\ln a}$ + C , a > 0, a ≠1
∫sinx dx	− cosx + C	∫cosx dx	sinx + C
$$\int_{}^{}{\frac{1}{\sin^{2}x}\text{\ dx}}$$	− ctgx + C, sinx ≠ 0	$$\int_{}^{}{\frac{1}{\cos^{2}x}\text{\ dx}}$$	tgx + C, cosx ≠ 0
$$\int_{}^{}{\frac{1}{{1 + \ x}^{2}}\text{\ dx}}$$	arctgx + C	$$\int_{}^{}{\frac{1}{\sqrt{1 - x^{2}}}\text{\ dx}}$$	arcsinx + C, −1 < x < 1
∫dx =  ∫x^0dx	x + C	$$\int_{}^{}{\sqrt{x}\text{\ dx}}$$	$\frac{2}{3}\ \sqrt{x^{3}} + \ $C
∫x dx	$\frac{x^{2}}{2}$ + C	∫x^2 dx	$\frac{x^{3}}{3}$ + C
$$\int_{}^{}{\frac{1}{x^{2}}\text{dx}}$$	$- \ \frac{1}{x} + \ $C	∫e^xdx	e^x + C
$$\int_{}^{}{\frac{1}{\sqrt{x}}\text{dx}}$$	2$\sqrt{x}$ + C	∫e^−xdx	−e^−x + C