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Identical expressions
(x+ two)*ln(x+ one)
(x plus 2) multiply by ln(x plus 1)
(x plus two) multiply by ln(x plus one)
(x+2)ln(x+1)
x+2lnx+1
(x+2)*ln(x+1)dx
Similar expressions
(x+2)*ln(x-1)
(x-2)*ln(x+1)

Integral of (x+2)*ln(x+1) dx

The solution

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

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(x + 2\right) \log{\left(x + 1 \right)} = x \log{\left(x + 1 \right)} + 2 \log{\left(x + 1 \right)}$
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. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 \log{\left(x + 1 \right)}\, dx = 2 \int \log{\left(x + 1 \right)}\, dx$
    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: $x - \log{\left(x + 1 \right)}$
    So, the result is: $- 2 x + 2 \left(x + 1\right) \log{\left(x + 1 \right)} - 2$
  The result is: $\frac{x^{2} \log{\left(x + 1 \right)}}{2} - \frac{x^{2}}{4} - \frac{3 x}{2} + 2 \left(x + 1\right) \log{\left(x + 1 \right)} - \frac{\log{\left(x + 1 \right)}}{2} - 2$
Method #2
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 $\operatorname{dv}{\left(x \right)} = x + 2$ .
  
  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:
      
      $\int 2\, dx = 2 x$
    The result is: $\frac{x^{2}}{2} + 2 x$
  Now evaluate the sub-integral.
2. Rewrite the integrand:
  
  $\frac{\frac{x^{2}}{2} + 2 x}{x + 1} = \frac{x}{2} + \frac{3}{2} - \frac{3}{2 \left(x + 1\right)}$
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:
    
    $\int \frac{3}{2}\, dx = \frac{3 x}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{3}{2 \left(x + 1\right)}\right)\, dx = - \frac{3 \int \frac{1}{x + 1}\, dx}{2}$
    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: $- \frac{3 \log{\left(x + 1 \right)}}{2}$
  The result is: $\frac{x^{2}}{4} + \frac{3 x}{2} - \frac{3 \log{\left(x + 1 \right)}}{2}$
Method #3
1. Rewrite the integrand:
  
  $\left(x + 2\right) \log{\left(x + 1 \right)} = x \log{\left(x + 1 \right)} + 2 \log{\left(x + 1 \right)}$
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. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 \log{\left(x + 1 \right)}\, dx = 2 \int \log{\left(x + 1 \right)}\, dx$
    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:
      
      $- x + \left(x + 1\right) \log{\left(x + 1 \right)} - 1$
    So, the result is: $- 2 x + 2 \left(x + 1\right) \log{\left(x + 1 \right)} - 2$
  The result is: $\frac{x^{2} \log{\left(x + 1 \right)}}{2} - \frac{x^{2}}{4} - \frac{3 x}{2} + 2 \left(x + 1\right) \log{\left(x + 1 \right)} - \frac{\log{\left(x + 1 \right)}}{2} - 2$
Now simplify:

$\frac{x^{2} \log{\left(x + 1 \right)}}{2} - \frac{x^{2}}{4} + 2 x \log{\left(x + 1 \right)} - \frac{3 x}{2} + \frac{3 \log{\left(x + 1 \right)}}{2} - 2$
Add the constant of integration:

$\frac{x^{2} \log{\left(x + 1 \right)}}{2} - \frac{x^{2}}{4} + 2 x \log{\left(x + 1 \right)} - \frac{3 x}{2} + \frac{3 \log{\left(x + 1 \right)}}{2} - 2+ \mathrm{constant}$

The answer is:

$\frac{x^{2} \log{\left(x + 1 \right)}}{2} - \frac{x^{2}}{4} + 2 x \log{\left(x + 1 \right)} - \frac{3 x}{2} + \frac{3 \log{\left(x + 1 \right)}}{2} - 2+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                     2    2                                  
 |                                  3*x   log(1 + x)   x    x *log(1 + x)                       
 | (x + 2)*log(x + 1) dx = -2 + C - --- - ---------- - -- + ------------- + 2*(1 + x)*log(1 + x)
 |                                   2        2        4          2                             
/

$$\int \left(x + 2\right) \log{\left(x + 1 \right)}\, dx = C + \frac{x^{2} \log{\left(x + 1 \right)}}{2} - \frac{x^{2}}{4} - \frac{3 x}{2} + 2 \left(x + 1\right) \log{\left(x + 1 \right)} - \frac{\log{\left(x + 1 \right)}}{2} - 2$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  9              15*log(3)
- - - 4*log(2) + ---------
  4                  2

$$- 4 \log{\left(2 \right)} - \frac{9}{4} + \frac{15 \log{\left(3 \right)}}{2}$$

  9              15*log(3)
- - - 4*log(2) + ---------
  4                  2

$$- 4 \log{\left(2 \right)} - \frac{9}{4} + \frac{15 \log{\left(3 \right)}}{2}$$

-9/4 - 4*log(2) + 15*log(3)/2

Numerical answer [src]

3.21700344277104

3.21700344277104

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (x+2)*ln(x+1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3