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Identical expressions
(one +x)*ln(one +x)
(1 plus x) multiply by ln(1 plus x)
(one plus x) multiply by ln(one plus x)
(1+x)ln(1+x)
1+xln1+x
(1+x)*ln(1+x)dx
Similar expressions
(1+x)*ln(1-x)
(1-x)*ln(1+x)

Integral of (1+x)*ln(1+x) dx

The solution

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  2                      
  /                      
 |                       
 |  (1 + x)*log(1 + x) dx
 |                       
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0

$$\int\limits_{0}^{2} \left(x + 1\right) \log{\left(x + 1 \right)}\, dx$$

Integral((1 + x)*log(1 + x), (x, 0, 2))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = x + 1$ .
  
  Then let $du = dx$ and substitute $du$ :
  
  $\int u \log{\left(u \right)}\, du$
  1. Let $u = \log{\left(u \right)}$ .
    
    Then let $du = \frac{du}{u}$ and substitute $du$ :
    
    $\int u e^{2 u}\, du$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
    Now substitute $u$ back in:
    
    $\frac{u^{2} \log{\left(u \right)}}{2} - \frac{u^{2}}{4}$
  Now substitute $u$ back in:
  
  $\frac{\left(x + 1\right)^{2} \log{\left(x + 1 \right)}}{2} - \frac{\left(x + 1\right)^{2}}{4}$
Method #2
1. Rewrite the integrand:
  
  $\left(x + 1\right) \log{\left(x + 1 \right)} = x \log{\left(x + 1 \right)} + \log{\left(x + 1 \right)}$
2. Integrate term-by-term:
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = \log{\left(x + 1 \right)}$ and let $\operatorname{dv}{\left(x \right)} = x$ .
    
    Then $\operatorname{du}{\left(x \right)} = \frac{1}{x + 1}$ .
    
    To find $v{\left(x \right)}$ :
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
    Now evaluate the sub-integral.
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{x^{2}}{2 \left(x + 1\right)}\, dx = \frac{\int \frac{x^{2}}{x + 1}\, dx}{2}$
    1. Rewrite the integrand:
      
      $\frac{x^{2}}{x + 1} = x - 1 + \frac{1}{x + 1}$
    2. Integrate term-by-term:
      
      The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-1\right)\, dx = - x$
      
      Let $u = x + 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x + 1 \right)}$
      
      The result is: $\frac{x^{2}}{2} - x + \log{\left(x + 1 \right)}$
    So, the result is: $\frac{x^{2}}{4} - \frac{x}{2} + \frac{\log{\left(x + 1 \right)}}{2}$
  1. There are multiple ways to do this integral.
    Method #1
    
    Let $u = x + 1$ .
    
    Then let $du = dx$ and substitute $du$ :
    
    $\int \log{\left(u \right)}\, du$
    
    Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(u \right)} = \log{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = 1$ .
    
    Then $\operatorname{du}{\left(u \right)} = \frac{1}{u}$ .
    
    To find $v{\left(u \right)}$ :
    
    The integral of a constant is the constant times the variable of integration:
    
    $\int 1\, du = u$
    
    Now evaluate the sub-integral.
    
    The integral of a constant is the constant times the variable of integration:
    
    $\int 1\, du = u$
    
    Now substitute $u$ back in:
    
    $- x + \left(x + 1\right) \log{\left(x + 1 \right)} - 1$
    Method #2
    
    Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = \log{\left(x + 1 \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$ .
    
    Then $\operatorname{du}{\left(x \right)} = \frac{1}{x + 1}$ .
    
    To find $v{\left(x \right)}$ :
    
    The integral of a constant is the constant times the variable of integration:
    
    $\int 1\, dx = x$
    
    Now evaluate the sub-integral.
    
    Rewrite the integrand:
    
    $\frac{x}{x + 1} = 1 - \frac{1}{x + 1}$
    
    Integrate term-by-term:
    
    The integral of a constant is the constant times the variable of integration:
    
    $\int 1\, dx = x$
    
    The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{x + 1}\right)\, dx = - \int \frac{1}{x + 1}\, dx$
    
    Let $u = x + 1$ .
    
    Then let $du = dx$ and substitute $du$ :
    
    $\int \frac{1}{u}\, du$
    
    The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    
    Now substitute $u$ back in:
    
    $\log{\left(x + 1 \right)}$
    
    So, the result is: $- \log{\left(x + 1 \right)}$
    
    The result is: $x - \log{\left(x + 1 \right)}$
  The result is: $\frac{x^{2} \log{\left(x + 1 \right)}}{2} - \frac{x^{2}}{4} - \frac{x}{2} + \left(x + 1\right) \log{\left(x + 1 \right)} - \frac{\log{\left(x + 1 \right)}}{2} - 1$
Method #3
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = \log{\left(x + 1 \right)}$ and let $\operatorname{dv}{\left(x \right)} = x + 1$ .
  
  Then $\operatorname{du}{\left(x \right)} = \frac{1}{x + 1}$ .
  
  To find $v{\left(x \right)}$ :
  1. Integrate term-by-term:
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
    The result is: $\frac{x^{2}}{2} + x$
  Now evaluate the sub-integral.
2. Rewrite the integrand:
  
  $\frac{\frac{x^{2}}{2} + x}{x + 1} = \frac{x}{2} + \frac{1}{2} - \frac{1}{2 \left(x + 1\right)}$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{x}{2}\, dx = \frac{\int x\, dx}{2}$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
    So, the result is: $\frac{x^{2}}{4}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{2}\, dx = \frac{x}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{2 \left(x + 1\right)}\right)\, dx = - \frac{\int \frac{1}{x + 1}\, dx}{2}$
    1. Let $u = x + 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x + 1 \right)}$
    So, the result is: $- \frac{\log{\left(x + 1 \right)}}{2}$
  The result is: $\frac{x^{2}}{4} + \frac{x}{2} - \frac{\log{\left(x + 1 \right)}}{2}$
Method #4
1. Rewrite the integrand:
  
  $\left(x + 1\right) \log{\left(x + 1 \right)} = x \log{\left(x + 1 \right)} + \log{\left(x + 1 \right)}$
2. Integrate term-by-term:
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = \log{\left(x + 1 \right)}$ and let $\operatorname{dv}{\left(x \right)} = x$ .
    
    Then $\operatorname{du}{\left(x \right)} = \frac{1}{x + 1}$ .
    
    To find $v{\left(x \right)}$ :
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
    Now evaluate the sub-integral.
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{x^{2}}{2 \left(x + 1\right)}\, dx = \frac{\int \frac{x^{2}}{x + 1}\, dx}{2}$
    1. Rewrite the integrand:
      
      $\frac{x^{2}}{x + 1} = x - 1 + \frac{1}{x + 1}$
    2. Integrate term-by-term:
      
      The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-1\right)\, dx = - x$
      
      Let $u = x + 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x + 1 \right)}$
      
      The result is: $\frac{x^{2}}{2} - x + \log{\left(x + 1 \right)}$
    So, the result is: $\frac{x^{2}}{4} - \frac{x}{2} + \frac{\log{\left(x + 1 \right)}}{2}$
  1. Let $u = x + 1$ .
    
    Then let $du = dx$ and substitute $du$ :
    
    $\int \log{\left(u \right)}\, du$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = \log{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = 1$ .
      
      Then $\operatorname{du}{\left(u \right)} = \frac{1}{u}$ .
      
      To find $v{\left(u \right)}$ :
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      Now evaluate the sub-integral.
    2. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
    Now substitute $u$ back in:
    
    $- x + \left(x + 1\right) \log{\left(x + 1 \right)} - 1$
  The result is: $\frac{x^{2} \log{\left(x + 1 \right)}}{2} - \frac{x^{2}}{4} - \frac{x}{2} + \left(x + 1\right) \log{\left(x + 1 \right)} - \frac{\log{\left(x + 1 \right)}}{2} - 1$
Now simplify:

$\frac{\left(x + 1\right)^{2} \left(2 \log{\left(x + 1 \right)} - 1\right)}{4}$
Add the constant of integration:

$\frac{\left(x + 1\right)^{2} \left(2 \log{\left(x + 1 \right)} - 1\right)}{4}+ \mathrm{constant}$

The answer is:

$\frac{\left(x + 1\right)^{2} \left(2 \log{\left(x + 1 \right)} - 1\right)}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                   2          2           
 |                             (1 + x)    (1 + x) *log(1 + x)
 | (1 + x)*log(1 + x) dx = C - -------- + -------------------
 |                                4                2         
/

$$\int \left(x + 1\right) \log{\left(x + 1 \right)}\, dx = C + \frac{\left(x + 1\right)^{2} \log{\left(x + 1 \right)}}{2} - \frac{\left(x + 1\right)^{2}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

     9*log(3)
-2 + --------
        2

$$-2 + \frac{9 \log{\left(3 \right)}}{2}$$

     9*log(3)
-2 + --------
        2

$$-2 + \frac{9 \log{\left(3 \right)}}{2}$$

-2 + 9*log(3)/2

Numerical answer [src]

2.94375529900649

2.94375529900649

Use the examples entering the upper and lower limits of integration.

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (1+x)*ln(1+x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2

Method #3

Method #4