(x­­^r)	rx^r-1
(c) c=const	0
(x)	1
(x^2)	2x
$$\left( \frac{1}{x} \right)$$	$$- \frac{1}{x^{2}}$$
$$\sqrt{x}$$	$$\frac{1}{2\sqrt{x}}$$
e^x	e^x
a^x	a^xlna
lnx	$$\frac{1}{x}$$
(x)	$$\frac{1}{\text{x\ lna}}$$
sinx	cosx
cosx	−sinx
tgx	$$\frac{1}{\operatorname{}x}$$
ctgx	$$- \frac{1}{\operatorname{}x}$$

[c*f(x)]^′ = c * f′(x)

[f(x)+f(x)]^′ = f^′(x) + g′(x)

[f(x)*g(x)]^′ = f^′(x) * g(x) + f(x) * g^′(x)

$$\left\lbrack \frac{f\left( x \right)}{g\left( x \right)} \right\rbrack^{'} = \frac{f^{'}\left( x \right)*g\left( x \right) - f\left( x \right)*g'(x)}{\left\lbrack g\left( x \right) \right\}^{2}}$$

((3x + 5)⁶)^′=6*(3x + 5)⁵*(3x + 5)^′=18*(3x + 5)⁵

(sin(2x + 1))^′=cos(2x + 1)*(2x + 1)^′=2cos(2x + 1)

sin²x=u^2′=2uu^′=(2sinx)^′=2sinxcosx

∫0dx	C
∫dx	x + C
∫x^rdx	$$\frac{1}{r + 1}x^{r + 1}$$
$$\int_{}^{}{\frac{1}{x}\text{dx}}$$	ln|x| + C
∫e^xdx	e^x + C
∫a^xdx	$$\frac{a^{x}}{\text{lna}} + C$$
∫sinxdx	−cosx + C
∫cosxdx	sinx + C
∫lnxdx	xlnx − x + C
∫tgxdx	−ln|cox| + C
∫ctgxdx	ln|sinx| + C
$$\int_{}^{}{\frac{1}{\cos x^{2}x}\text{dx}}$$	tgx + C
$$\int_{}^{}\frac{1}{\operatorname{}x}\text{dx}$$	−ctgx + C

$$\int_{}^{}\begin{matrix} \mathbf{f}\left( \mathbf{ax\ + \ b} \right)^{\mathbf{n}}\mathbf{dx\ =}\frac{\mathbf{1}}{\mathbf{a*}\left( \mathbf{n + 1} \right)}\mathbf{F}\left( \mathbf{ax\ + \ b} \right)^{\mathbf{n + 1}}\mathbf{+ \ C\ \ \ \ \ \ } \\ \mathbf{a \neq \ 0} \\ \end{matrix}$$

jak nie ma potegi to po prostu dwie caleczki

$$\int_{}^{}\frac{1}{2x + 3}dx = \frac{1}{2}\ln\left| 2x + 3 \right| + C$$

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

$$\int_{}^{}{\frac{2x}{x^{2} + 4}dx = \ln\left( x^{2} + 4 \right) + C\ \ \ bo\ \ \left( x^{2} + 4 \right)^{'} = 2x}$$

∫uv′dx=uv−∫u^′vdx   calkowanie przez czesci

$$\int_{}^{}{xsin2xdx} = \begin{Bmatrix} u = x & v^{'} = sin2x \\ u^{'} = 1 & v = - \frac{1}{2}cos2x \\ \end{Bmatrix} = x\left( - \frac{1}{2}cos2x) \right) - \int_{}^{}{1*\left( - \frac{1}{2}cos2x \right)\text{dx}}$$

∫f(x)dx=∫f(g(t))g^′(t)dt_{↑ t = y(x)} podstawienie

$$\int_{}^{}{\frac{1}{a^{x}} = - \frac{1}{a^{x}\text{lna}}} + C$$

sin3x = sinx(3 − 4x) sin2x = 2sinxcosx

cos3x = cosx(4x − 3) cos2x = x  − x