Integral de 1/2*ln(|x+1|) dx
Solución
Solución detallada
La integral del producto de una función por una constante es la constante por la integral de esta función:
∫ log ( ∣ x + 1 ∣ ) 2 d x = ∫ log ( ∣ x + 1 ∣ ) d x 2 \int \frac{\log{\left(\left|{x + 1}\right| \right)}}{2}\, dx = \frac{\int \log{\left(\left|{x + 1}\right| \right)}\, dx}{2} ∫ 2 l o g ( ∣ x + 1 ∣ ) d x = 2 ∫ l o g ( ∣ x + 1 ∣ ) d x
Usamos la integración por partes:
∫ u dv = u v − ∫ v du \int \operatorname{u} \operatorname{dv}
= \operatorname{u}\operatorname{v} -
\int \operatorname{v} \operatorname{du} ∫ u dv = u v − ∫ v du
que u ( x ) = log ( ∣ x + 1 ∣ ) u{\left(x \right)} = \log{\left(\left|{x + 1}\right| \right)} u ( x ) = log ( ∣ x + 1 ∣ ) y que dv ( x ) = 1 \operatorname{dv}{\left(x \right)} = 1 dv ( x ) = 1 .
Entonces du ( x ) = ( ( re ( x ) + 1 ) d d x re ( x ) + im ( x ) d d x im ( x ) ) sign ( x + 1 ) ( x + 1 ) ∣ x + 1 ∣ \operatorname{du}{\left(x \right)} = \frac{\left(\left(\operatorname{re}{\left(x\right)} + 1\right) \frac{d}{d x} \operatorname{re}{\left(x\right)} + \operatorname{im}{\left(x\right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}\right) \operatorname{sign}{\left(x + 1 \right)}}{\left(x + 1\right) \left|{x + 1}\right|} du ( x ) = ( x + 1 ) ∣ x + 1 ∣ ( ( re ( x ) + 1 ) d x d re ( x ) + im ( x ) d x d im ( x ) ) sign ( x + 1 ) .
Para buscar v ( x ) v{\left(x \right)} v ( x ) :
La integral de las constantes tienen esta constante multiplicada por la variable de integración:
∫ 1 d x = x \int 1\, dx = x ∫ 1 d x = x
Ahora resolvemos podintegral.
Hay varias maneras de calcular esta integral.
Método #1
Vuelva a escribir el integrando:
x ( ( re ( x ) + 1 ) d d x re ( x ) + im ( x ) d d x im ( x ) ) sign ( x + 1 ) ( x + 1 ) ∣ x + 1 ∣ = x re ( x ) sign ( x + 1 ) d d x re ( x ) + x im ( x ) sign ( x + 1 ) d d x im ( x ) + x sign ( x + 1 ) d d x re ( x ) x ∣ x + 1 ∣ + ∣ x + 1 ∣ \frac{x \left(\left(\operatorname{re}{\left(x\right)} + 1\right) \frac{d}{d x} \operatorname{re}{\left(x\right)} + \operatorname{im}{\left(x\right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}\right) \operatorname{sign}{\left(x + 1 \right)}}{\left(x + 1\right) \left|{x + 1}\right|} = \frac{x \operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)} + x \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)} + x \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{x \left|{x + 1}\right| + \left|{x + 1}\right|} ( x + 1 ) ∣ x + 1 ∣ x ( ( re ( x ) + 1 ) d x d re ( x ) + im ( x ) d x d im ( x ) ) sign ( x + 1 ) = x ∣ x + 1 ∣ + ∣ x + 1 ∣ x re ( x ) sign ( x + 1 ) d x d re ( x ) + x im ( x ) sign ( x + 1 ) d x d im ( x ) + x sign ( x + 1 ) d x d re ( x )
Vuelva a escribir el integrando:
x re ( x ) sign ( x + 1 ) d d x re ( x ) + x im ( x ) sign ( x + 1 ) d d x im ( x ) + x sign ( x + 1 ) d d x re ( x ) x ∣ x + 1 ∣ + ∣ x + 1 ∣ = x re ( x ) sign ( x + 1 ) d d x re ( x ) x ∣ x + 1 ∣ + ∣ x + 1 ∣ + x im ( x ) sign ( x + 1 ) d d x im ( x ) x ∣ x + 1 ∣ + ∣ x + 1 ∣ + x sign ( x + 1 ) d d x re ( x ) x ∣ x + 1 ∣ + ∣ x + 1 ∣ \frac{x \operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)} + x \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)} + x \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{x \left|{x + 1}\right| + \left|{x + 1}\right|} = \frac{x \operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{x \left|{x + 1}\right| + \left|{x + 1}\right|} + \frac{x \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{x \left|{x + 1}\right| + \left|{x + 1}\right|} + \frac{x \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{x \left|{x + 1}\right| + \left|{x + 1}\right|} x ∣ x + 1 ∣ + ∣ x + 1 ∣ x re ( x ) sign ( x + 1 ) d x d re ( x ) + x im ( x ) sign ( x + 1 ) d x d im ( x ) + x sign ( x + 1 ) d x d re ( x ) = x ∣ x + 1 ∣ + ∣ x + 1 ∣ x re ( x ) sign ( x + 1 ) d x d re ( x ) + x ∣ x + 1 ∣ + ∣ x + 1 ∣ x im ( x ) sign ( x + 1 ) d x d im ( x ) + x ∣ x + 1 ∣ + ∣ x + 1 ∣ x sign ( x + 1 ) d x d re ( x )
Integramos término a término:
No puedo encontrar los pasos en la búsqueda de esta integral.
Pero la integral
∫ x re ( x ) sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x \int \frac{x \operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx ∫ ( x + 1 ) ∣ x + 1 ∣ x re ( x ) sign ( x + 1 ) d x d re ( x ) d x
No puedo encontrar los pasos en la búsqueda de esta integral.
Pero la integral
∫ x im ( x ) sign ( x + 1 ) d d x im ( x ) ( x + 1 ) ∣ x + 1 ∣ d x \int \frac{x \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx ∫ ( x + 1 ) ∣ x + 1 ∣ x im ( x ) sign ( x + 1 ) d x d im ( x ) d x
No puedo encontrar los pasos en la búsqueda de esta integral.
Pero la integral
∫ x sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x \int \frac{x \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx ∫ ( x + 1 ) ∣ x + 1 ∣ x sign ( x + 1 ) d x d re ( x ) d x
El resultado es: ∫ x sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x + ∫ x re ( x ) sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x + ∫ x im ( x ) sign ( x + 1 ) d d x im ( x ) ( x + 1 ) ∣ x + 1 ∣ d x \int \frac{x \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx + \int \frac{x \operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx + \int \frac{x \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx ∫ ( x + 1 ) ∣ x + 1 ∣ x sign ( x + 1 ) d x d re ( x ) d x + ∫ ( x + 1 ) ∣ x + 1 ∣ x re ( x ) sign ( x + 1 ) d x d re ( x ) d x + ∫ ( x + 1 ) ∣ x + 1 ∣ x im ( x ) sign ( x + 1 ) d x d im ( x ) d x
Método #2
Vuelva a escribir el integrando:
x ( ( re ( x ) + 1 ) d d x re ( x ) + im ( x ) d d x im ( x ) ) sign ( x + 1 ) ( x + 1 ) ∣ x + 1 ∣ = x re ( x ) sign ( x + 1 ) d d x re ( x ) x ∣ x + 1 ∣ + ∣ x + 1 ∣ + x im ( x ) sign ( x + 1 ) d d x im ( x ) x ∣ x + 1 ∣ + ∣ x + 1 ∣ + x sign ( x + 1 ) d d x re ( x ) x ∣ x + 1 ∣ + ∣ x + 1 ∣ \frac{x \left(\left(\operatorname{re}{\left(x\right)} + 1\right) \frac{d}{d x} \operatorname{re}{\left(x\right)} + \operatorname{im}{\left(x\right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}\right) \operatorname{sign}{\left(x + 1 \right)}}{\left(x + 1\right) \left|{x + 1}\right|} = \frac{x \operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{x \left|{x + 1}\right| + \left|{x + 1}\right|} + \frac{x \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{x \left|{x + 1}\right| + \left|{x + 1}\right|} + \frac{x \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{x \left|{x + 1}\right| + \left|{x + 1}\right|} ( x + 1 ) ∣ x + 1 ∣ x ( ( re ( x ) + 1 ) d x d re ( x ) + im ( x ) d x d im ( x ) ) sign ( x + 1 ) = x ∣ x + 1 ∣ + ∣ x + 1 ∣ x re ( x ) sign ( x + 1 ) d x d re ( x ) + x ∣ x + 1 ∣ + ∣ x + 1 ∣ x im ( x ) sign ( x + 1 ) d x d im ( x ) + x ∣ x + 1 ∣ + ∣ x + 1 ∣ x sign ( x + 1 ) d x d re ( x )
Integramos término a término:
No puedo encontrar los pasos en la búsqueda de esta integral.
Pero la integral
∫ x re ( x ) sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x \int \frac{x \operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx ∫ ( x + 1 ) ∣ x + 1 ∣ x re ( x ) sign ( x + 1 ) d x d re ( x ) d x
No puedo encontrar los pasos en la búsqueda de esta integral.
Pero la integral
∫ x im ( x ) sign ( x + 1 ) d d x im ( x ) ( x + 1 ) ∣ x + 1 ∣ d x \int \frac{x \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx ∫ ( x + 1 ) ∣ x + 1 ∣ x im ( x ) sign ( x + 1 ) d x d im ( x ) d x
No puedo encontrar los pasos en la búsqueda de esta integral.
Pero la integral
∫ x sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x \int \frac{x \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx ∫ ( x + 1 ) ∣ x + 1 ∣ x sign ( x + 1 ) d x d re ( x ) d x
El resultado es: ∫ x sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x + ∫ x re ( x ) sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x + ∫ x im ( x ) sign ( x + 1 ) d d x im ( x ) ( x + 1 ) ∣ x + 1 ∣ d x \int \frac{x \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx + \int \frac{x \operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx + \int \frac{x \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx ∫ ( x + 1 ) ∣ x + 1 ∣ x sign ( x + 1 ) d x d re ( x ) d x + ∫ ( x + 1 ) ∣ x + 1 ∣ x re ( x ) sign ( x + 1 ) d x d re ( x ) d x + ∫ ( x + 1 ) ∣ x + 1 ∣ x im ( x ) sign ( x + 1 ) d x d im ( x ) d x
Por lo tanto, el resultado es: x log ( ∣ x + 1 ∣ ) 2 − ∫ x sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 − ∫ x re ( x ) sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 − ∫ x im ( x ) sign ( x + 1 ) d d x im ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 \frac{x \log{\left(\left|{x + 1}\right| \right)}}{2} - \frac{\int \frac{x \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2} - \frac{\int \frac{x \operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2} - \frac{\int \frac{x \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2} 2 x l o g ( ∣ x + 1 ∣ ) − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x sign ( x + 1 ) d x d re ( x ) d x − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x re ( x ) sign ( x + 1 ) d x d re ( x ) d x − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x im ( x ) sign ( x + 1 ) d x d im ( x ) d x
Ahora simplificar:
x log ( ∣ x + 1 ∣ ) 2 − ∫ x sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 − ∫ x re ( x ) sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 − ∫ x im ( x ) sign ( x + 1 ) d d x im ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 \frac{x \log{\left(\left|{x + 1}\right| \right)}}{2} - \frac{\int \frac{x \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2} - \frac{\int \frac{x \operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2} - \frac{\int \frac{x \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2} 2 x l o g ( ∣ x + 1 ∣ ) − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x sign ( x + 1 ) d x d re ( x ) d x − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x re ( x ) sign ( x + 1 ) d x d re ( x ) d x − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x im ( x ) sign ( x + 1 ) d x d im ( x ) d x
Añadimos la constante de integración:
x log ( ∣ x + 1 ∣ ) 2 − ∫ x sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 − ∫ x re ( x ) sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 − ∫ x im ( x ) sign ( x + 1 ) d d x im ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 + c o n s t a n t \frac{x \log{\left(\left|{x + 1}\right| \right)}}{2} - \frac{\int \frac{x \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2} - \frac{\int \frac{x \operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2} - \frac{\int \frac{x \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2}+ \mathrm{constant} 2 x l o g ( ∣ x + 1 ∣ ) − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x sign ( x + 1 ) d x d re ( x ) d x − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x re ( x ) sign ( x + 1 ) d x d re ( x ) d x − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x im ( x ) sign ( x + 1 ) d x d im ( x ) d x + constant
Respuesta:
x log ( ∣ x + 1 ∣ ) 2 − ∫ x sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 − ∫ x re ( x ) sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 − ∫ x im ( x ) sign ( x + 1 ) d d x im ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 + c o n s t a n t \frac{x \log{\left(\left|{x + 1}\right| \right)}}{2} - \frac{\int \frac{x \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2} - \frac{\int \frac{x \operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2} - \frac{\int \frac{x \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2}+ \mathrm{constant} 2 x l o g ( ∣ x + 1 ∣ ) − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x sign ( x + 1 ) d x d re ( x ) d x − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x re ( x ) sign ( x + 1 ) d x d re ( x ) d x − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x im ( x ) sign ( x + 1 ) d x d im ( x ) d x + constant
Respuesta (Indefinida)
[src]
/ / /
| | |
| d | d | d
| x*--(re(x))*sign(1 + x) | x*--(im(x))*im(x)*sign(1 + x) | x*--(re(x))*re(x)*sign(1 + x)
| dx | dx | dx
| ----------------------- dx | ----------------------------- dx | ----------------------------- dx
/ | (1 + x)*|1 + x| | (1 + x)*|1 + x| | (1 + x)*|1 + x|
| | | |
| log(|x + 1|) / / / x*log(|x + 1|)
| ------------ dx = C - ----------------------------- - ----------------------------------- - ----------------------------------- + --------------
| 2 2 2 2 2
|
/
∫ log ( ∣ x + 1 ∣ ) 2 d x = C + x log ( ∣ x + 1 ∣ ) 2 − ∫ x sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 − ∫ x re ( x ) sign ( x + 1 ) d d x re ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 − ∫ x im ( x ) sign ( x + 1 ) d d x im ( x ) ( x + 1 ) ∣ x + 1 ∣ d x 2 \int \frac{\log{\left(\left|{x + 1}\right| \right)}}{2}\, dx = C + \frac{x \log{\left(\left|{x + 1}\right| \right)}}{2} - \frac{\int \frac{x \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2} - \frac{\int \frac{x \operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2} - \frac{\int \frac{x \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x + 1 \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left(x + 1\right) \left|{x + 1}\right|}\, dx}{2} ∫ 2 log ( ∣ x + 1 ∣ ) d x = C + 2 x log ( ∣ x + 1 ∣ ) − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x sign ( x + 1 ) d x d re ( x ) d x − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x re ( x ) sign ( x + 1 ) d x d re ( x ) d x − 2 ∫ ( x + 1 ) ∣ x + 1 ∣ x im ( x ) sign ( x + 1 ) d x d im ( x ) d x
Gráfica
0.00 1.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.0 0.5
− 1 2 + log ( 2 ) - \frac{1}{2} + \log{\left(2 \right)} − 2 1 + log ( 2 )
=
− 1 2 + log ( 2 ) - \frac{1}{2} + \log{\left(2 \right)} − 2 1 + log ( 2 )
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.