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Integrale indefinito

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Svolgimento integrale indefinito .
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INTEGRALE INDEFINITO

.. , ( 3x + 4x + 4x + x + 2x + 1 )*dx / [ ( x + 1 )*( x + 1 ) ] }
= , ( 3x + 4x + 4x + x + 2x + 1 )*dx / * ( x + 1 )*( x - x + 1 )*( x + 1 ) ] }

Grazie alla linearita degli integrali :

= * A*dx / ( x + 1 ) + + * ( B*x + C )*dx / ( x - x + 1 ) ] +
.. + * ( D*x + E )*dx / ( x + 1 ) + + * ( F*x + G )*dx / ( x + 1 ) +

Applicando il principio d'identita dei polinomi determiniamo i valori dei parametri
A, B, C, D, E, F, G :

.. A*( x - x + 1 )*( x + 1 ) + ( B*x + C )*( x + 1 )*( x + 1 ) +
+ ( D*x + E )*( x + 1 )*( x - x + 1 )*( x + 1 ) + ( F*x + G )*( x + 1 )*( x - x + 1 ) =
= 3x + 4x + 4x + x + 2x + 1

.. ( A + B + D )*x + ( - A + B + C + E )*x + ( 3A + 2B + C + D + F )*x +
+ ( - 2A + 2B + 2C + D + E + G )*x + ( 3A + B + 2C + E )*x +
+ ( - A + B + C + D + F )*x + ( A + C + E + G ) =
= 3x + 4x + 4x + x + 2x + 1

Da cui segue :

{ A + B + D = 0
{ - A + B + C + E = 3
{ 3A + 2B + C + D + F = 4
{ - 2A + 2B + 2C + D + E + G = 4

{ 3A + B + 2C + E = 1
{ - A + B + C + D + F = 2
{ A + C + E + G = 1

Ovvero :

{ A = - 1/4
{ B = 3
{ C = - 1
{ D = - 11/4
{ E = 3/4
{ F = 5/2
{ G = 3/2

= - ( 1/4 )** dx / ( x + 1 ) + + * ( 3x - 1 )*dx / ( x - x + 1 ) ] +
.. + ( 1/4 )** ( 3 - 11x )*dx / ( x + 1 ) + + ( 1/2 )** ( 5x + 3 )*dx / ( x + 1 ) +

= - ( 1/4 )** dx / ( x + 1 ) + + 3** x*dx / ( x - x + 1 ) ] +
.. - * dx / ( x - x + 1 ) + + ( 3/4 )** dx / ( x + 1 ) + +
.. - ( 11/4 )** x*dx / ( x + 1 ) + + ( 1/2 )** ( 5x + 3 )*dx / ( x + 1 ) +

= - ( 1/4 )** dx / ( x + 1 ) + + ( 3/2 )** 2x*dx / ( x - x + 1 ) ] +
.. - * dx / ( x - x + 1 ) + + ( 3/4 )** dx / ( x + 1 ) + +
.. - ( 11/8 )** 2x*dx / ( x + 1 ) + + ( 1/2 )** ( 5x + 3 )*dx / ( x + 1 ) +

= - ( 1/4 )** dx / ( x + 1 ) + + ( 3/2 )** ( 2x - 1 + 1 )*dx / ( x - x + 1 ) ] +
.. - * dx / ( x - x + 1 ) + + ( 3/4 )** dx / ( x + 1 ) + +
.. - ( 11/8 )** 2x*dx / ( x + 1 ) + + ( 1/2 )** ( 5x + 3 )*dx / ( x + 1 ) ]


= - ( 1/4 )** dx / ( x + 1 ) + + ( 3/2 )** ( 2x - 1 )*dx / ( x - x + 1 ) ] +
.. + ( 1/2 )** dx / ( x - x + 1 ) + + ( 3/4 )** dx / ( x + 1 ) + +
.. - ( 11/8 )** 2x*dx / ( x + 1 ) + + ( 1/2 )** ( 5x + 3 )*dx / ( x + 1 ) +

Ricordando che in generale :

* f ' ( x ) * dx / f ( x ) + = LN| f ( x ) | + c ........ con c
, f ' ( x )*dx / * a + ( f ( x ) ) + - = ( 1 / a )*ATAN* f ( x ) / a + + c ... con ( a 0, c )

= - ( 1/4 )*LN| x + 1 | + ( 3/2 )*LN| x - x + 1 | +
.. + ( 1/2 )** dx / ( x - x + 1 ) ] + ( 3/4 )*ATAN( x ) +
.. - ( 11/8 )*LN| x + 1 | + ( 1/2 )*[ ( 5x + 3 )*dx / ( x + 1 ) ]

Grazie alle proprieta dei logaritmi :

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 1/2 )** dx / ( x - x + 1/4 + 3/4 ) + + ( 1/2 )*[ ( 5x + 3 )*dx / ( x + 1 ) ]

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 1/2 )*, dx / * ( x - 1/2 ) + 3/4 + - + ( 1/2 )*[ ( 5x + 3 )*dx / ( x + 1 ) ]

Ricordando nuovamente che in generale :
, f ' ( x )*dx / * a + ( f ( x ) ) + - = ( 1 / a )*ATAN* f ( x ) / a + + c ... con ( a 0, c )

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 1/2 )*( 2/3 )*ATAN* ( x - 1/2 ) / ( 3/2 ) + + ( 1/2 )** ( 5x + 3 )*dx / ( x + 1 ) +


= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) + + ( 1/2 )** ( 5x + 3 )*dx / ( x + 1 ) +

Sostituzione :
t = ATAN( x ) ...... x = TAN( t ) ...... dx = [ TAN( t ) + 1 ]*dt = dt / COS( t )

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) ] +
.. + ( 1/2 )*, * 5*TAN( t ) + 3 +** TAN( t ) + 1 +*dt / * TAN( t ) + 1 + -

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) ] +
.. + ( 1/2 )*, * 5*TAN( t ) + 3 +*dt / * 1 / COS( t ) + -

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) ] +
.. + ( 1/2 )*, * 3*COS( t ) + 5*SIN( t )*COS( t ) +*dt -

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) ] +
.. + ( 3/2 )** COS( t )*dt ] + ( 5/2 )** COS( t )*SIN( t )*dt +

Ricordando che in generale :
* f ' ( t )*f ( t )^( a )*dt + = f ( t )^( a + 1 ) / ( a + 1 ) + c ... con ( a -1 , c )

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) + + ( 3/2 )** COS( t )*dt + + ( 5/4 )*SIN( t )


Ricordando la formula di duplicazione del coseno :
COS( 2t ) = 1 - 2*COS( t )

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) + + ( 3/2 )*, * 1 - COS( 2t ) ]*dt / 2 } +
.. + ( 5/4 )*SIN( t )

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) + + ( 3/4 )** dt + + ( 3/4 )** COS( 2t )*dt +
.. + ( 5/4 )*SIN( t )

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) + + ( 3/4 )*t + ( 3/8 )** 2*COS( 2t )*dt +
.. + ( 5/4 )*SIN( t )

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) ] + ( 3/4 )*t + ( 3/8 )*SIN( 2t ) +
.. + ( 5/4 )*SIN( t ) + c

Ricordando la formula di duplicazione del seno :
SIN( 2t ) = 2*SIN( t )*COS( t )

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) ] + ( 3/4 )*t + ( 3/4 )*SIN( t )*COS( t ) +
.. + ( 5/4 )*SIN( t ) + c

Ricordando che avevamo posto ...... t = ATAN( x ) ......


= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) ] + ( 3/4 )*ATAN( x ) +
.. + ( 3/4 )*SIN[ ATAN( x ) ]*COS[ ATAN( x ) ] + ( 5/4 )*SIN[ ATAN( x ) ] + c

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) ] + ( 3/4 )*ATAN( x ) +
.. + ( 3/4 )** x / ( 1 + x ) +** 1 / ( 1 + x ) + + ( 5/4 )** x / ( 1 + x ) + + c

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) ] + ( 3/4 )*ATAN( x ) +
.. + ( 3/4 )*x / ( 1 + x ) + ( 5/4 )*x / ( 1 + x ) + c

= ( 1/8 )*{ 6*ATAN( x ) - LN[ ( x + 1 ) ] + 12*LN( x - x + 1 ) - 11*LN( x + 1 ) } +
.. + ( 3 / 3 )*ATAN* ( 3 / 3 )*( 2x - 1 ) ] + ( 3/4 )*ATAN( x ) +
.. + ( 1/4 )*x*( 5x + 3 ) / ( 1 + x ) + c

_______________________________ _____________________________________
POST SCRIPTUM :
Per convincersi delle seguenti identita :
SIN* ATAN( x ) + = x / ( 1 + x )
COS* ATAN( x ) = 1 / ( 1 + x )
e sufficiente disegnare un triangolo ABC di base AB = 1, altezza AC = x, retto in A e con angolo in B pari ad
Per il teorema di Pitagora ... CB = ( 1 + x ) .
Per evidenti proprieta trigonometriche ... x = AB*TAN( ) = TAN( )
.................... ............................. ...... SIN( ) = x / ( 1 + x ) .
Da .. x = TAN( ) ... ricaviamo che ... = ATAN( x ) .
Segue che ... SIN* ATAN( x ) + = x / ( 1 + x ) .



Stesso ragionamento lo si puo fare per ricavare l'espressione algebrica di
COS[ ATAN( x ) ] .

Nota che tali identita sono ben poste per il fatto che valgono pure per x < 0 ... l'arcotangente infatti e una
funzioni dispari !
_______________________________ _____________________________________


Ciao ciao

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