¿Cómo integro [math] \ frac {4 \ ln \ left | x \ right |} {x [1 + (\ ln \ left | x \ right |) ^ 2]} [/ math]?

Evaluar:

[matemáticas] \ large \ displaystyle I = \ large \ displaystyle \ int \ frac {4 \ ln \ left | x \ right |} {x [1 + (\ ln \ left | x \ right |) ^ 2]} \ , dx [/ math]

Sustituir: [matemáticas] \ large \ displaystyle \ ln \ left | x \ right | = \ large \ displaystyle t [/ math]

[math] \ implica \ large \ displaystyle \ frac {1} {x} dx = \ large \ displaystyle dt [/ math]

[matemáticas] \ implica \ large \ displaystyle I = \ large \ displaystyle 2 \ int \ frac {2t} {1 + t ^ 2} \, dt [/ math]

Sustituir (de nuevo) : [math] \ large \ displaystyle (1 + t ^ 2) = \ large \ displaystyle y [/ math]

[math] \ implica \ large \ displaystyle 2t dt = \ large \ displaystyle dy [/ math]

[math] \ implica \ large \ displaystyle I = \ large \ displaystyle 2 \ int \ frac {1} {y} \, dy [/ math]

[matemáticas] \ implica \ large \ displaystyle I = \ large \ displaystyle 2 \ ln \ left | y \ right | + c [/ matemáticas]

Convertir de nuevo a la variable original, es decir, [math] \ large \ displaystyle x [/ math] a través de [math] \ large \ displaystyle t [/ math]

[matemáticas] \ implica \ large \ displaystyle I = \ large \ displaystyle 2 \ ln \ left | 1 + t ^ 2 \ derecha | + c [/ matemáticas]

[matemáticas] \ implica \ large \ displaystyle \ boxed {\ boxed {I = \ large \ displaystyle 2 \ ln \ left | 1 + (\ ln \ left | x \ right | \ right |) ^ 2 + c}} [/ math]

¡Hecho!

[matemática] {\ enorme {\ enorme {\ displaystyle \ ddot \ smile}}} [/ matemática]

Primero, dejemos que [math] t = \ ln x. [/ Math] Luego [math] dt = \ frac {1} {x} dx [/ math] y tenemos eso

[matemáticas] \ int \ frac {4 \ ln x} {x (1 + \ ln ^ 2 x)} dx = \ int \ frac {4t} {1 + t ^ 2} dt. [/ math]

Deje [math] u = t ^ 2, [/ math] luego [math] du = 2tdt [/ math] y tenemos

[matemáticas] \ int \ frac {4t} {1 + t ^ 2} dt = \ frac {2du} {1 + u} = 2 \ ln (| 1 + u |) + C = 2 \ ln (1 + t ^ 2) = 2 \ ln (1 + \ ln ^ 2 x) + C [/ matemáticas]