Sr Examen

Integral de ln(1+|x|) dx

Límites de integración:

interior superior
v

Gráfico:

interior superior

Definida a trozos:

Solución

Ha introducido [src]
  2                
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 |  log(1 + |x|) dx
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-2                 
22log(x+1)dx\int\limits_{-2}^{2} \log{\left(\left|{x}\right| + 1 \right)}\, dx
Integral(log(1 + |x|), (x, -2, 2))
Solución detallada
  1. Usamos la integración por partes:

    udv=uvvdu\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

    que u(x)=log(x+1)u{\left(x \right)} = \log{\left(\left|{x}\right| + 1 \right)} y que dv(x)=1\operatorname{dv}{\left(x \right)} = 1.

    Entonces du(x)=(re(x)ddxre(x)+im(x)ddxim(x))sign(x)x(x+1)\operatorname{du}{\left(x \right)} = \frac{\left(\operatorname{re}{\left(x\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 \right)}}{x \left(\left|{x}\right| + 1\right)}.

    Para buscar v(x)v{\left(x \right)}:

    1. La integral de las constantes tienen esta constante multiplicada por la variable de integración:

      1dx=x\int 1\, dx = x

    Ahora resolvemos podintegral.

  2. Hay varias maneras de calcular esta integral.

    Método #1

    1. Vuelva a escribir el integrando:

      (re(x)ddxre(x)+im(x)ddxim(x))sign(x)x+1=re(x)sign(x)ddxre(x)+im(x)sign(x)ddxim(x)x+1\frac{\left(\operatorname{re}{\left(x\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 \right)}}{\left|{x}\right| + 1} = \frac{\operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)} + \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left|{x}\right| + 1}

    2. Vuelva a escribir el integrando:

      re(x)sign(x)ddxre(x)+im(x)sign(x)ddxim(x)x+1=re(x)sign(x)ddxre(x)x+1+im(x)sign(x)ddxim(x)x+1\frac{\operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)} + \operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left|{x}\right| + 1} = \frac{\operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left|{x}\right| + 1} + \frac{\operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left|{x}\right| + 1}

    3. Integramos término a término:

      1. No puedo encontrar los pasos en la búsqueda de esta integral.

        Pero la integral

        re(x)sign(x)ddxre(x)x+1dx\int \frac{\operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left|{x}\right| + 1}\, dx

      1. No puedo encontrar los pasos en la búsqueda de esta integral.

        Pero la integral

        im(x)sign(x)ddxim(x)x+1dx\int \frac{\operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left|{x}\right| + 1}\, dx

      El resultado es: re(x)sign(x)ddxre(x)x+1dx+im(x)sign(x)ddxim(x)x+1dx\int \frac{\operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left|{x}\right| + 1}\, dx + \int \frac{\operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left|{x}\right| + 1}\, dx

    Método #2

    1. Vuelva a escribir el integrando:

      (re(x)ddxre(x)+im(x)ddxim(x))sign(x)x+1=re(x)sign(x)ddxre(x)x+1+im(x)sign(x)ddxim(x)x+1\frac{\left(\operatorname{re}{\left(x\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 \right)}}{\left|{x}\right| + 1} = \frac{\operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left|{x}\right| + 1} + \frac{\operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left|{x}\right| + 1}

    2. Integramos término a término:

      1. No puedo encontrar los pasos en la búsqueda de esta integral.

        Pero la integral

        re(x)sign(x)ddxre(x)x+1dx\int \frac{\operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left|{x}\right| + 1}\, dx

      1. No puedo encontrar los pasos en la búsqueda de esta integral.

        Pero la integral

        im(x)sign(x)ddxim(x)x+1dx\int \frac{\operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left|{x}\right| + 1}\, dx

      El resultado es: re(x)sign(x)ddxre(x)x+1dx+im(x)sign(x)ddxim(x)x+1dx\int \frac{\operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left|{x}\right| + 1}\, dx + \int \frac{\operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left|{x}\right| + 1}\, dx

  3. Añadimos la constante de integración:

    xlog(x+1)re(x)sign(x)ddxre(x)x+1dxim(x)sign(x)ddxim(x)x+1dx+constantx \log{\left(\left|{x}\right| + 1 \right)} - \int \frac{\operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left|{x}\right| + 1}\, dx - \int \frac{\operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left|{x}\right| + 1}\, dx+ \mathrm{constant}


Respuesta:

xlog(x+1)re(x)sign(x)ddxre(x)x+1dxim(x)sign(x)ddxim(x)x+1dx+constantx \log{\left(\left|{x}\right| + 1 \right)} - \int \frac{\operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left|{x}\right| + 1}\, dx - \int \frac{\operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left|{x}\right| + 1}\, dx+ \mathrm{constant}

Respuesta (Indefinida) [src]
                           /                               /                                           
                          |                               |                                            
                          | d                             | d                                          
  /                       | --(im(x))*im(x)*sign(x)       | --(re(x))*re(x)*sign(x)                    
 |                        | dx                            | dx                                         
 | log(1 + |x|) dx = C -  | ----------------------- dx -  | ----------------------- dx + x*log(1 + |x|)
 |                        |         1 + |x|               |         1 + |x|                            
/                         |                               |                                            
                         /                               /                                             
log(x+1)dx=C+xlog(x+1)re(x)sign(x)ddxre(x)x+1dxim(x)sign(x)ddxim(x)x+1dx\int \log{\left(\left|{x}\right| + 1 \right)}\, dx = C + x \log{\left(\left|{x}\right| + 1 \right)} - \int \frac{\operatorname{re}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{re}{\left(x\right)}}{\left|{x}\right| + 1}\, dx - \int \frac{\operatorname{im}{\left(x\right)} \operatorname{sign}{\left(x \right)} \frac{d}{d x} \operatorname{im}{\left(x\right)}}{\left|{x}\right| + 1}\, dx
Respuesta [src]
-4 + 6*log(3)
4+6log(3)-4 + 6 \log{\left(3 \right)}
=
=
-4 + 6*log(3)
4+6log(3)-4 + 6 \log{\left(3 \right)}
-4 + 6*log(3)
Respuesta numérica [src]
2.59127214330217
2.59127214330217

    Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.