Mostrando entradas con la etiqueta matemàtiques-càlcul-integral. Mostrar todas las entradas
Mostrando entradas con la etiqueta matemàtiques-càlcul-integral. Mostrar todas las entradas

martes, 21 de julio de 2020

producte integral tangent

TAN_{x}[f(x)] = ( 1/(1+( f(x) )^{2}) )·d_{x}[f(x)]


TAN_{x}[tan(x)] = 1


f(x) [o( tan(x) )o] g(x) = ∫ [ TAN_{x}[f(x)]·TAN_{x}[g(x)] ] TAN[x]


f(x) [o( tan(x) )o] tan(x) = f(x)


f(x) [o( tan(x) )o] ( f(x) )^{[o( tan(x) )o](-1)} = tan(x)

jueves, 30 de abril de 2020

integració per parts

∫ [ f(x)·d_{x}[g(x)]·( mx^{(1+(-1)(1/m))} ) ] d[x^{(1/m)}] = ...
... f(x)·g(x)+(-1)·∫ [ g(x)·d_{x}[f(x)]·( mx^{(1+(-1)(1/m))} ) ] d[x^{(1/m)}]


∫ [ f(x)·d_{x^{(1/m)}}[g(x)] ] d[x^{(1/m)}] = ...
... f(x)·g(x)+(-1)·∫ [ g(x)·d_{x^{(1/m)}}[f(x)] ] d[x^{(1/m)}]

sábado, 28 de marzo de 2020

derivació irracional

d_{x^{(1/n)}}[x^{(1/n)}] = 1


d_{x^{(1/n)}}[x^{(m/n)}] = mx^{( (m+(-1))/n )}

derivació irracional

d_{x^{(1/n)}}[g( f(x^{(1/n)}) )] = d_{f(x^{(1/n)})}[g( f(x^{(1/n)}) )]·d_{x^{(1/n)}}[f(x^{(1/n)})]

integral irracional

∫ [ ( ax+bx^{(1/2)} )^{n} ] d[x] = ∫ [ ( ax+bx^{(1/2)} )^{n}·( 2x^{(1/2)} ) ] d[x^{(1/2)}] = ...
... ( 1/(n+1) )( ax+bx^{(1/2)} )^{(n+1)} [o( x^{(1/2)} )o] ...
... ( 1/(2a) )·ln( 2ax^{(1/2)}+b ) [o( x^{(1/2)} )o] x


∫ [ ( ax+bx^{(2/3)}+cx^{(1/3)} )^{n} ] d[x] = ...
... ∫ [ ( ax+bx^{(2/3)}+cx^{(1/3)} )^{n}·( 3x^{(2/3)} ) ] d[x^{(1/3)}] = ...
... ( 1/(n+1) )( ax+bx^{(2/3)}+cx^{(1/3)} )^{(n+1)} [o( x^{(1/3)} )o] ...
... ln( 3ax^{(2/3)}+2bx^{(1/3)}+c ) [o( x^{(1/3)} )o] ...
... ( 1/(6a) )·ln( 6ax^{(1/3)}+2b ) [o( x^{(1/3)} )o] x

jueves, 26 de marzo de 2020

producte integral sinus-y-cosinus elíptic

sin[n](x) = ...
... ∑ ( e^{(2/(n+1))·pi·i} )^{k_{1}...k_{n}}·( 1/((2·k_{1}...k_{n})+1)! )·x^{( (2k_{1}...k_{n})+1 )}
cos[n](x) = ...
... ∑ ( e^{(2/(n+1))·pi·i} )^{k_{1}...k_{n}}·( 1/(2·k_{1}...k_{n})! )·x^{(2k_{1}...k_{n})}


d_{x}[sin[n](x)] = cos[n](x)
d_{x}[cos[n](x)] = e^{(2/(n+1))·pi·i}·sin[n](x)
d_{xx}^{2}[sin[n](x)] = e^{(2/(n+1))·pi·i}·sin[n](x)
d_{xx}^{2}[cos[n](x)] = e^{(2/(n+1))·pi·i}·cos[n](x)


( cos[n](x) )^{(n+1)} + ( sin[n](x) )^{(n+1)} = n^{(n+1)}
d_{x}[cos[n](x)]^{(n+1)} + d_{x}[sin[n](x)]^{(n+1)} = n^{(n+1)}
d_{xx}^{2}[cos[n](x)]^{(n+1)} + d_{xx}^{2}[sin[n](x)]^{(n+1)} = n^{(n+1)}


f(x) [o( sin[n](x) )o] sin[n](x) = f(x)


S[n]_{x}[f(x)] = ( d_{x}[f(x)]/( n^{(n+1)}+(-1)·( f(x) )^{(n+1)} )^{(1/(n+1))} )


f(x) [o( cos[n](x) )o] cos[n](x) = f(x)


C[n]_{x}[f(x)] = e^{( (2n)/(n+1) )·pi·i}·( d_{x}[f(x)]/( n^{(n+1)}+(-1)·( f(x) )^{(n+1)})^{(1/(n+1))} )


teorema:
sin[n]( f(x) [o(x)o] g(x) ) = sin[n](f(x)) [o( sin[n](x) )o] sin[n](g(x))
cos[n]( f(x) [o(x)o] g(x) ) = cos[n](f(x)) [o( cos[n](x) )o] cos[n](g(x))

integral de producte integral logarítmic

∫ [ ( f(x) )^{n} ] L[x] = ...
... ln( (1/(n+1))·( f(x) )^{(n+1)} ) [o( ln(x) )o] ( ln( f(x) ) )^{[o( ln(x) )o)](-1)}


∫ [ ( ax^{2}+bx )^{n} ] L[x] = ...
... ln( (1/(n+1))·( ax^{2}+bx )^{(n+1)} ) [o( ln(x) )o] ln( ( 1/(2a) )·ln(2ax+b) )


∫ [ ( ax^{3}+bx^{2}+cx )^{n} ] L[x] = ...
... ln( (1/(n+1))·( ax^{3}+bx^{2}+cx )^{(n+1)} ) [o( ln(x) )o] ...
... ln( ln(3ax^{2}+2bx+c) ) [o( ln(x) )o] ln( ( 1/(6a) )·ln(6ax+2b) )

producte integral logarítmic

f(x) [o( ln(x) )o] g(x) = ∫ [ L_{x}[f(x)]·L_{x}[g(x)] ] L[x]


f(x) [o( ln(x) )o] ln(x) = f(x)


f(x) [o( ln(x) )o] ( f(x) )^{[o( ln(x) )o](-1)} = ln(x)


teorema:
ln( f(x) [o(x)o] g(x) ) = ln(f(x)) [o( ln(x) )o] ln(g(x))


demostració
L_{x}[ ln( f(x) [o(x)o] g(x) ) ] = L_{x}[ ln(f(x)) [o( ln(x) )o] ln(g(x)) ]

derivació logarítmica

L_{x}[f(x)] = e^{f(x)}·d_{x}[f(x)]


L_{x}[c] = 0


L_{x}[x^{n}] = e^{x^{n}}·nx^{(n+(-1))}


L_{x}[ln(x)] = 1

lunes, 23 de marzo de 2020

integral de producte integral exponencial


∫ [ e^{( f(x) )^{n}} ] D[x] = e^{ e^{( f(x) )^{n}} } [v(x)v] ( e^{( f(x) )^{n}} )^{[v(x)v](-1)}

integral de producte integral exponencial


∫ [ e^{f(x)} ] D[x] = e^{ e^{f(x)} } [v(x)v] ( e^{f(x)} )^{[v(x)v](-1)}

integral de producte integral exponencial


∫ [ ( f(x) )^{n} ] D[x] = e^{(1/(n+1))·( f(x) )^{(n+1)}} [v(x)v] ( e^{f(x)} )^{[v(x)v](-1)}


∫ [ ( ax^{2}+bx )^{n} ] D[x] = ...
... e^{(1/(n+1))·( ax^{2}+bx )^{(n+1)}} [v(x)v] e^{(1/(2a))·ln(2ax+b)}


∫ [ ( ax^{3}+bx^{2}+cx )^{n} ] D[x] = ...
... e^{(1/(n+1))·( ax^{3}+bx^{2}+cx )^{(n+1)}} [v(x)v] ...
... e^{ln(3ax^{2}+2bx+c)} [v(x)v] e^{(1/(6a))·ln(6ax+2b)}

producte integral exponencial

f(x) [v(x)v] g(x) = ∫ [ D_{x}[f(x)]·D_{x}[g(x)] ] D[x]


( f(x) [v(x)v] g(x) ) [v(x)v] h(x) = f(x) [v(x)v] ( g(x) [v(x)v] h(x) )
∫ [ ( D_{x}[f(x)]·D_{x}[g(x)] )·D_{x}[h(x)] ] D[x] = ∫ [ D_{x}[f(x)]·( D_{x}[g(x)]·D_{x}[h(x)] ) ] D[x]


e^{x} [v(x)v] g(x) = ∫ [ D_{x}[g(x)] ] D[x] = g(x)


f(x) [v(x)v] ( f(x) )^{[v(x)v](-1)} = e^{x}


teorema:
e^{f(x) [o(x)o] g(x)} = e^{f(x)} [v(x)v] e^{g(x)}


demostració:
D_{x}[ e^{f(x) [o(x)o] g(x)} ] = D_{x}[ e^{f(x)} [v(x)v] e^{g(x)} ]


proposició:
e^{x [o(x)o] f(x)} = e^{x} [v(x)v] e^{f(x)}

derivació y integració exponencial

D_{x}[f(x)] = lim [h-->0][ (f(x+h)+(-1)·f(x))/(h·f(x)) ] = ( d_{x}[f(x)]/f(x) )


D_{x}[c] = 0


D_{x}[x^{n}] = (n/x)


D_{x}[e^{x}] = 1


D_{x}[a^{x}] = ln(a)


D_{x}[ln(x)] = ( 1/ln(x) )·(1/x)


D_{x}[f(x)+g(x)] = D_{x}[f(x)] + D_{x}[g(x)]


D_{x}[a·f(x)] = D_{x}[f(x)]


D_{x}[f(x)·g(x)] = D_{x}[f(x)]·g(x) + f(x)·D_{x}[g(x)]


D_{x}[ax^{2}+bx+c] = (2ax+b)/(ax^{2}+bx)


∫ [ D_{x}[f(x)] ] d[x] = ln( ∫ [ D_{x}[f(x)] ] D[x] )


e^{ ∫ [ D_{x}[f(x)] ] d[x] } = ∫ [ D_{x}[f(x)] ] D[x]


e^{∫ d[x]} = ∫ D[x] = e^{x}


e^{∫ [(n/x)] d[x]} = ∫ (n/x) D[x]
e^{ln(x^{n})} = x^{n}


e^{∫ [ ln(a) ] d[x]} = ∫ [ ln(a) ] D[x]
e^{ln(a)·x} = a^{x}


e^{∫ [ ( 1/ln(x) )·(1/x) ] d[x]} = ∫ [ ( 1/ln(x) )·(1/x) ] D[x]
e^{ln(ln(x))} = ln(x)


D_{x}[f(g(x))] = d_{g(x)}[f(g(x))]·d_{x}[g(x)]·(1/f(g(x)))


e^{∫ [ x^{n} ] d[x]} = ∫ [ x^{n} ] D[x] = e^{(1/(n+1))·x^{(n+1)}}
D_{x}[ e^{(1/(n+1))·x^{(n+1)}} ] = x^{n}

sábado, 22 de febrero de 2020

integrals

∫ [ ( ln(x) )^{n} ] d[x] = (1/(n+1))·( ln(x) )^{n+1} [o(x)o] (1/2)·x^{2}


∫ [ ( ln( f(x) ) )^{n} ] d[x] = ...
... (1/(n+1))·( ln( f(x) ) )^{n+1} [o(x)o] ∫ [ f(x) ] d[x] [o(x)o] ( f(x) )^{[o(x)o](-1)}


∫ [ ( ln( ax^{2}+bx ) )^{n} ] d[x] = ...
... (1/(n+1))·( ln( ax^{2}+bx ) )^{n+1} [o(x)o] ( a·(1/3)·x^{3}+b·(1/2)·x^{2} ) [o(x)o] (1/2a)·ln( 2ax+b )

lunes, 20 de enero de 2020

integrals exponencials


∫ [e^{( f(x) )^{n}}] d[x] = e^{( f(x) )^{n}} [o(x)o] ( ( f(x) )^{n} )^{[o(x)o](-1)}


∫ [e^{( ax )^{n}}] d[x] = ...
...e^{( ax )^{n}} [o(x)o] (1/a^{2})·(1/n)·( 1/((-n)+2) )·( ( ax )^{(-n)+2} )


∫ [e^{( ax^{2}+bx )^{n}}] d[x] = ...
...e^{( ax^{2}+bx )^{n}} [o(x)o]...
...(1/n)·( 1/((-n)+2) )·( ( ax^{2}+bx )^{(-n)+2} ) [o(x)o] (1/2a)·(-1)·( 2ax+b )^{(-1)}


∫ [e^{( ax^{3}+bx^{2}+cx )^{n}}] d[x] = ...
...e^{( ax^{3}+bx^{2}+cx )^{n}} [o(x)o]...
...(1/n)·( 1/((-n)+2) )·( ( ax^{3}+bx^{2}+cx )^{(-n)+2} ) [o(x)o] ...
...(-1)·( 3ax^{2}+2bx+c )^{(-1)} [o(x)o] (1/6a)·ln(6ax+b)


∫ [e^{( ax^{4}+bx^{3}+cx^{2}+dx )^{n}}] d[x] = ...
...e^{( ax^{4}+bx^{3}+cx^{2}+dx )^{n}} [o(x)o]...
...(1/n)·( 1/((-n)+2) )·( ( ax^{4}+bx^{3}+cx^{2}+dx )^{(-n)+2} ) [o(x)o] ...
...(-1)·( 4ax^{3}+3bx^{2}+2cx+d )^{(-1)} [o(x)o] ln(12ax^{2}+6bx+2c) [o(x)o] (1/24a)·ln(24ax+6b)

integrals potencials


∫ [( f(x) )^{n}] d[x] = (1/(n+1))( f(x) )^{(n+1)} [o(x)o] ( f(x) )^{[o(x)o](-1)}


∫ [ ( ax^{2}+bx )^{n} ] d[x] = (1/(n+1))( ax^{2}+bx )^{(n+1)} [o(x)o] (1/2a)·ln(2ax+b)


∫ [ ( ax^{3}+bx^{2}+cx )^{n} ] d[x] = ...
...(1/(n+1))( ax^{3}+bx^{2}+cx )^{(n+1)} [o(x)o] ( ln(3ax^{2}+2bx+c) [o(x)o] (1/6a)·ln(6ax+2b) )


∫ [ ( ax^{4}+bx^{3}+cx^{2}+dx )^{n} ] d[x] = ...
...(1/(n+1))( ax^{4}+bx^{3}+cx^{2}+dx )^{(n+1)} [o(x)o] ...
...( ln(4ax^{3}+3bx^{2}+2cx+d) [o(x)o] ln(12ax^{2}+6bx+2c) [o(x)o] (1/24a)·ln(24ax+6b) )

integral de serie geométrica

∫ [ e^{x}·( (x^{(n+1)}+(-1))/(x+(-1)) ) ] d[x] = e^{x}+∑ ( x^{(k+1)}·er_{m;k+1}(x) )


∫ [ e^{x}·( (x^{(n+1)}+(-1))/(x+(-1)) ) ] d[x]= ∫ [ e^{x}+∑ ( e^{x}·x^{k} ) ] d[x]


∫ [ e^{(-x)}·( (x^{(n+1)}+(-1))/(x+(-1)) ) ] d[x] = (-1)·( e^{(-x)}+∑ ( (-x)^{(k+1)}·er_{m;k+1}(-x) ) )

martes, 24 de diciembre de 2019

tangent y cotangent integral

tan[o(x)o](x) = ∫ [ (-1)·(cos(x)/sin(x)) ] d[x] = ∫ [ (-1)cotan(x) ] d[x]


cotan[o(x)o](x) = ∫ [ (-1)^{(-1)}·(sin(x)/cos(x)) ] d[x] = ∫ [ (-1)^{(-1)}·tan(x) ] d[x]


d_{x}[ tan[o(x)o](x) ] = (-1)·cotan(x)


d_{x}[ cotan[o(x)o](x) ] = (-1)^{(-1)}·tan(x)


d_{x}[ ( tan[o(x)o](x) )^{[o(x)o]n} ] = ( (-1)·cotan(x) )^{n}


d_{x}[ ( cotan[o(x)o](x) )^{[o(x)o]n} ] = ( (-1)^{(-1)}·tan(x) )^{n}