Problema de integrales?

Question

Integral de (ln4x/ln2x) dx es? Si pudieran con algo de proced. tengo idea pero quisiera consultarlo de todas formas ...

supongo que lo harán por integración por partes pero no tengo la integral de logaritmos ...

llegué a x + Ln(2) * Integral [dx / Ln(2x)]
pero no tengo mucha idea para lo que falta

llegué a x + Ln(2) * Integral [dx / Ln(2x)]
pero no tengo mucha idea para lo que falta

Answer 1

Respuesta: x + Ln(2) * Integral [dx / Ln(2x)]
___________
Como: Ln 4x / Ln 2x = Ln (2*2x) / Ln 2x = (Ln 2 + Ln 2x) / Ln 2x =
= 1 + [ Ln(2) / Ln(2x) ] entonces:
Integral [Ln(4x) / Ln(2x) ] dx = x + Ln(2) * Integral [dx / Ln(2x)]
___________
Esta última integral no tiene una primitiva. Es más, hay una función que se utiliza en la Teoría Analítica de Números llamada "logaritmo integral" (Li(x)" que no tiene integral sino una aproximación numérica y que se define por:

Li(x) = Integral de "2" a "x" de [dt / Ln(t)]
___________
Se concluye que si esta integral se calcula para extremos definidos mediante cálculos numéricos aproximados, entonces no tiene una primitiva.
...

Answer 2

x+1/2*ln(2)*LogaritmoIntegral(2x)

Answer 3

Te recomiendo el libro Calculo de Larson

Answer 4

Integration Techniques
I. Basic antidifferentiation rules you should know:
Z xn dx =
1
n + 1xn+1 if n 6= −1 Z 1
x
dx = ln|x|
Z ex dx = ex Z sin(x) dx = −cos(x)
Z cos(x) dx = sin(x) Z tan(x) dx = ln|sec(x)| = −ln |cos(x)|
Z cot(x) dx = ln|sin(x)| = −ln |csc(x)| Z sec2(x) dx = tan(x)
Z csc2(x) dx = −cot(x) Z sec(x) tan(x) dx = sec(x)
Z csc(x) cot(x) dx = −csc(x) Z 1
x2 + 1
(x) dx = arctan(x)
Z 1
p1 − x2
dx = arcsin(x)
II. If your integrand does not appear on the above list, it may be possible to simplify the integrand
by expanding it.
Example 1. Z 􀀀x2 + 12
dx = Z x4 + 2x2 + 1 dx
Example 2. Z x+ 1
x2 dx = Z x
x2 +
1
x2 dx = Z 1
x
+ x−2 dx
III. If it is not possible to simplify your integrand, try a substitution. Rules of thumb for deciding
what to choose for u when using substitution:
1. If an expression appears raised to a power or under a root, let u = that expression.
Example. To find Z p2x + 5 dx let u = 2x+ 5.
2. If one part of the integrand is the derivative of a second part of the integrand, let u = the
second part.
Example. To find Z 3x2 sin 􀀀x3 dx, let u = x3.
OVER
IV. If substitution does not work, try integration by parts: Z u dv = uv − Z v du
Rule of thumb for deciding what to choose for u when using integration by parts: ILATE (Inverse
trig functions, Logarithms, Algebraic expressions like powers of x, Trig functions, Exponential
functions). If the integrand contains an inverse trig function, let u be that piece of the integrand.
If there is no inverse function in the integrand but there is a logarithm, let u be the logarithm.
If there are no inverse functions or logs but there is an algebraic expression (e.g. a power of
x), let u be the algebraic expression, etc. Don’t forget that you may have to integrate by parts
more than once.
Example. To find Z 4x ln(x) dx let u = ln(x) and dv = 4x dx. Then u = ln(x) ) du =
1
x
dx and
dv = 4x dx ) v = Z 4x dx = 2x2, so Z 4x ln(x) dx
| u{dzv }
= 2x2 ln(x)
| {z } uv
−Z 2x dx |v{dzu}
= 2x2 ln(x) − x2
V. If the integrand contains a term of the form pa2 − u2, pa2 − u2, or pa2 + u2, a trigonometric
substitution may be helpful.
1. For integrands involving pa2 − u2 let u = a sin(t). E.g., for Z 1
x2p25 − x2
dx,
let x = 5 sin(t).
2. For integrands involving pu2 − a2 let u = a sec(t). E.g., for Z 1
px2 − 9
dx, let x = 3 sec(t).
3. For integrands involving pa2 + u2 let u = a tan(t). E.g., for Z p4x2 + 9
x4 dx,
let 2x = 3 tan(t).
VI. If the integrand is a proper rational function, try a partial fraction decomposition.
Example. To find Z x2 − 2x+ 1
x4 + x2 dx, use partial fractions:
x2 − 2x + 1
x4 + x2 = x2 − 2x + 1
x2 (x2 + 1)
= A
x
+ B
x2 + Cx + D
x2 + 1 )
x2 − 2x + 1 = A
x
+ B
x2 + Cx + D
x2 + 1 􀀀x4 + x2 = Ax 􀀀x2 + 1+ B 􀀀x2 + 1+ (Cx + D) x2
= (A + C) x3 + (B + D) x2 + Ax + B )
8>
><>>:
A + C = 0
B + D = 1
A = −2
B = 1
) A = −2, B = 1, C = 2, D = 0 )
Z x2 − 2x+ 1
x4 + x2 dx = Z −2
x
+
1
x2 +
2x
x2 + 1 dx = −2 ln|x| − x−1 + ln􀀀x2 + 1+ c


Espero que sea algo de lo que buscas. Se ve mejor en la direccion que te proporciono. ok. por que aqui me cambia algunos simbolos, pero bueno espero te sirva.

Answer 5

buena pregunta, por favor inicia la respuesta, a ver donde llegas