if y = log b x then b y = x {\displaystyle {\text{if}}\ y={\text{log}}_{b}\ x\ {\text{then}}\ b^{y}=x}
log b b = 1 and log b 1 = 0 {\displaystyle {\text{log}}_{b}\ b=1\ {\text{and}}\ {\text{log}}_{b}\ 1=0}
log b b x = x {\displaystyle {\text{log}}_{b}\ b^{x}=x}
b log b x = x {\displaystyle b^{{\text{log}}_{b}^{\ }x}=x}
log a x = log b x log b a {\displaystyle {\text{log}}_{a}x={\frac {{\text{log}}_{b}x}{{\text{log}}_{b}a}}}
log b ( x r ) = r log b x {\displaystyle {\text{log}}_{b}\left(x^{r}\right)=r\ {\text{log}}_{b}x}
log b ( x y ) = log b x + log b y {\displaystyle {\text{log}}_{b}(xy)={\text{log}}_{b}x+{\text{log}}_{b}y}
log b ( x y ) = log b x − log b y {\displaystyle {\text{log}}_{b}\left({\frac {x}{y}}\right)={\text{log}}_{b}x-{\text{log}}_{b}y}
x 2 − a 2 = ( x + a ) ( x − a ) {\displaystyle x^{2}-a^{2}=(x+a)(x-a)}
x 2 + 2 a x + a 2 = ( x + a ) 2 {\displaystyle x^{2}+2ax+a^{2}=(x+a)^{2}}
x 2 − 2 a x − a 2 = ( x − a ) 2 {\displaystyle x^{2}-2ax-a^{2}=(x-a)^{2}}
x 2 + ( a + b ) x + a b = ( x + a ) ( x + b ) {\displaystyle x^{2}+(a+b)x+ab=(x+a)(x+b)}
x 2 + 3 a x 2 + 3 a 2 x + a 3 = ( x + a ) 3 {\displaystyle x^{2}+3ax^{2}+3a^{2}x+a^{3}=(x+a)^{3}}
x 3 + a 3 = ( x + a ) ( x 2 − a x a 2 ) {\displaystyle x^{3}+a^{3}=(x+a)(x^{2}-ax_{a}^{2})}
x 3 − a 3 = ( x − a ) ( x 2 + a x + a 2 ) {\displaystyle x^{3}-a^{3}=(x-a)(x^{2}+ax+a^{2})}
x 2 n − a 2 n = ( x n − a n ) ( x n + a n ) {\displaystyle x^{2n}-a^{2n}=(x^{n}-a^{n})(x^{n}+a^{n})}
| a | = { a , if a ≥ 0 − a , if a < 0 {\displaystyle \left|a\right|={\begin{cases}a,{\mbox{if}}\ a\geq 0\\-a,{\mbox{if}}\ a<0\end{cases}}}
| a | = | − a | {\displaystyle \left|a\right|=\left|-a\right|}
| a | ≥ 0 {\displaystyle \left|a\right|\geq 0}
| a b | = | a | | b | {\displaystyle \left|ab\right|=\left|a\right|\left|b\right|}
| a b | = | a | | b | {\displaystyle \left|{\frac {a}{b}}\right|={\frac {\left|a\right|}{\left|b\right|}}}
| a + b | ≤ | a | + | b | {\displaystyle \left|a+b\right|\leq \left|a\right|+\left|b\right|}
∫ a b f ( x ) d x = lim n → ∞ ∑ k = 1 n f ( x k ) Δ x {\displaystyle \int _{a}^{b}f(x)dx=\lim _{n\to \infty }\sum _{k=1}^{n}f(x_{k})\Delta x}
where Δ x = b − a n and x k = a + k Δ x {\displaystyle {\mbox{where}}\ \Delta x={\frac {b-a}{n}}\ {\mbox{and}}\ x_{k}=a+k\Delta x}
∫ a b f ( x ) d x = [ F ( x ) ] a b = F ( b ) − F ( a ) {\displaystyle \int _{a}^{b}f(x)dx=[F(x)]_{a}^{b}=F(b)-F(a)}
∫ a b c f ( x ) d x = c ∫ a b f ( x ) d x {\displaystyle \int _{a}^{b}cf(x)dx=c\int _{a}^{b}f(x)dx}
∫ a b f ( x ) ± g ( x ) d x = ∫ a b f ( x ) d x ± ∫ a b g ( x ) d x {\displaystyle \int _{a}^{b}f(x)\pm g(x)dx=\int _{a}^{b}f(x)dx\pm \int _{a}^{b}g(x)dx}
∫ a b f ( x ) d x = 0 and ∫ a b f ( x ) d x = − ∫ b a f ( x ) d x {\displaystyle \int _{a}^{b}f(x)dx=0\ {\mbox{and}}\ \int _{a}^{b}f(x)dx=-\int _{b}^{a}f(x)dx}
∫ a b f ( x ) d x + ∫ b c f ( x ) d x = ∫ a c f ( x ) d x {\displaystyle \int _{a}^{b}f(x)dx+\int _{b}^{c}f(x)dx=\int _{a}^{c}f(x)dx}
∫ k d x = k x + C {\displaystyle \int k\ dx=kx+C}
∫ x n d x = 1 n + 1 x n + 1 + C , n ≠ − 1 {\displaystyle \int x^{n}dx={\frac {1}{n+1}}x^{n+1}+C,n\neq -1}
∫ x − 1 d x = ∫ 1 x d x = ln | x | + C {\displaystyle \int x^{-1}dx=\int {\frac {1}{x}}dx={\mbox{ln}}\left|x\right|+C}
∫ 1 a x + b d x = 1 a ln | a x + b | + C {\displaystyle \int {\frac {1}{ax+b}}dx={\frac {1}{a}}{\mbox{ln}}\left|ax+b\right|+C}
∫ ln ( x ) d x = x ln ( x ) − x + C {\displaystyle \int {\mbox{ln}}(x)\ dx=x\ {\mbox{ln}}(x)-x+C}
∫ e x d x = e x + C {\displaystyle \int e^{x}\ dx=e^{x}+C}
∫ cos x d x = sin x + C {\displaystyle \int \cos x\ dx=\sin x+C}
∫ sin x d x = − cos x + C {\displaystyle \int \sin x\ dx=-\cos x+C}
∫ sec 2 x d x = tan x + C {\displaystyle \int \sec ^{2}x\ dx=\tan x+C}
∫ sec x tan x d x = sec x + C {\displaystyle \int \sec x\tan x\ dx=\sec x+C}
∫ csc x cot x d x = − csc x + C {\displaystyle \int \csc x\cot x\ dx=-\csc x+C}
∫ csc 2 x d x = − cot x + C {\displaystyle \int \csc ^{2}x\ dx=-\cot x+C}
∫ tan x d x = ln | sec x | + C {\displaystyle \int \tan x\ dx=\ln \left|\sec x\right|+C}
∫ sec x d x = ln | sec x + tan x | + C {\displaystyle \int \sec x\ dx=\ln \left|\sec x+\tan x\right|+C}
∫ 1 a 2 + u 2 d x = 1 a tan − 1 ( u a ) + C {\displaystyle \int {\frac {1}{a^{2}+u^{2}}}dx={\frac {1}{a}}\tan ^{-1}\left({\frac {u}{a}}\right)+C}
∫ 1 a 2 − u 2 d x = sin − 1 ( u a ) + C {\displaystyle \int {\frac {1}{\sqrt {a^{2}-u^{2}}}}dx=\sin ^{-1}\left({\frac {u}{a}}\right)+C}
Left-hand and right-hand rectangle approximations
L n = Δ x ∑ k = 0 n − 1 f ( x k ) R n = Δ x ∑ k = 1 n f ( x k ) {\displaystyle L_{n}=\Delta x\sum _{k=0}^{n-1}f(x_{k})\qquad \qquad R_{n}=\Delta x\sum _{k=1}^{n}f(x_{k})}
Midpoint Rule
M n = Δ x ∑ k = 0 n − 1 f ( x k + k k + 1 2 ) {\displaystyle M_{n}=\Delta x\sum _{k=0}^{n-1}f({\frac {x_{k}+k_{k+1}}{2}})}
Trapezoid Rule
T n = Δ x 2 ( f ( x 0 ) + 2 f ( x 1 ) + 2 f ( x 2 ) + ⋯ + f ( x n ) ) {\displaystyle T_{n}={\frac {\Delta x}{2}}(f(x_{0})+2f(x_{1})+2f(x_{2})+\cdots +f(x_{n}))}
S n = Δ x 3 ( f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + ⋯ + 2 f ( x n − 2 ) + 4 f ( x n − 1 + f ( x n ) ) {\displaystyle S_{n}={\frac {\Delta x}{3}}(f(x_{0})+4f(x_{1})+2f(x_{2})+\cdots +2f(x_{n-2})+4f(x_{n-1}+f(x_{n}))}
∫ a b f ( g ( x ) ) g ′ ( x ) d x = ∫ g ( a ) g ( b ) f ( u ) d u {\displaystyle \int _{a}^{b}f(g(x))g'(x)dx=\int _{g(a)}^{g(b)}f(u)du}
Where u = g ( x ) {\displaystyle u=g(x)} and d u = g ′ ( x ) d x {\displaystyle du=g'(x)dx}
∫ u d v = u v − ∫ v d u where v = ∫ d v {\displaystyle \int u\ dv=uv-\int v\ du\ {\mbox{where}}\ v=\int dv}
or
∫ f ( x ) g ′ ( x ) d x = f ( x ) g ( x ) − ∫ f ′ ( x ) g ( x ) d x {\displaystyle \int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx}
d d x ( f ( x ) ) = f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h {\displaystyle {\frac {d}{dx}}\left(f(x)\right)=f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}
( c f ( x ) ) ′ = c ( f ′ ( x ) ) {\displaystyle (cf(x))'=c(f'(x))}
( f ( x ) ± g ( x ) ) ′ = f ′ ( x ) ± g ′ ( x ) {\displaystyle (f(x)\pm g(x))'=f'(x)\pm g'(x)}
d d x ( c ) = 0 {\displaystyle {\frac {d}{dx}}\left(c\right)=0}
If f {\displaystyle f} is differentiable on the interval ( a , b {\displaystyle a,b} ) and continuous at the end points lhere exists a c {\displaystyle c} in ( a , b {\displaystyle a,b} ) such that
f ′ ( c ) = f ( b ) − f ( a ) b − a {\displaystyle f'\left(c\right)={\frac {f(b)-f(a)}{b-a}}}
( f ( x ) g ( x ) ) ′ = f ( x ) ′ g ( x ) + f ( x ) g ( x ) ′ {\displaystyle (f(x)g(x))'=f(x)'g(x)+f(x)g(x)'}
d d x ( f ( x ) d ( x ) ) = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) [ g ( x ) ] 2 {\displaystyle {\frac {d}{dx}}\left({\frac {f(x)}{d(x)}}\right)={\frac {f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}}}
![{\displaystyle \int _{a}^{b}f(x)dx=[F(x)]_{a}^{b}=F(b)-F(a)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f35df72551cd47d6fd7dfff2e19a83ee8f56d1a)
![{\displaystyle {\frac {d}{dx}}\left({\frac {f(x)}{d(x)}}\right)={\frac {f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/077ab594825ec834fd46ad64ed7b475f868ed369)