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Integral of d{x}:
Integral of x*e^(3*x)*dx
Integral of 1/(2x-1)
Integral of x/cos^2x
Integral of log(x+1)/(x+1)
Graphing y =:
log(x+1)/(x+1)
Derivative of:
log(x+1)/(x+1)
Identical expressions
log(x+ one)/(x+ one)
logarithm of (x plus 1) divide by (x plus 1)
logarithm of (x plus one) divide by (x plus one)
logx+1/x+1
log(x+1) divide by (x+1)
log(x+1)/(x+1)dx
Similar expressions
log(x-1)/(x+1)
log(x+1)/(x-1)
What you mean?
log(x + 1)/(x + 1)
log(x + 1)/x + 1
log(x + 1)/(x + 1)
log(x + 1)/x + 1
log(x + 1)/(x + 1)
log(x + 1)/x + 1
log(x + 1)/x + 1

You entered:

log(x+1)/(x+1)

What you mean?

log(x + 1)/(x + 1)

log(x + 1)/x + 1

log(x + 1)/(x + 1)

log(x + 1)/x + 1

log(x + 1)/(x + 1)

log(x + 1)/x + 1

log(x + 1)/x + 1

Integral of log(x+1)/(x+1) dx

The solution

You have entered [src]

  1              
  /              
 |               
 |  log(x + 1)   
 |  ---------- dx
 |    x + 1      
 |               
/                
0

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

Simplify

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = \log{\left(x + 1 \right)}$ .
  
  Then let $du = \frac{dx}{x + 1}$ and substitute $du$ :
  
  $\int u\, du$
  1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int u\, du = \frac{u^{2}}{2}$
  Now substitute $u$ back in:
  
  $\frac{\log{\left(x + 1 \right)}^{2}}{2}$
Method #2
1. Let $u = x + 1$ .
  
  Then let $du = dx$ and substitute $du$ :
  
  $\int \frac{\log{\left(u \right)}}{u}\, du$
  1. Let $u = \frac{1}{u}$ .
    
    Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$ :
    
    $\int \frac{\log{\left(\frac{1}{u} \right)}}{u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\log{\left(\frac{1}{u} \right)}}{u}\right)\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}}{u}\, du$
      
      Let $u = \log{\left(\frac{1}{u} \right)}$ .
      
      Then let $du = - \frac{du}{u}$ and substitute $- du$ :
      
      $\int u\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u\right)\, du = - \int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}$
      
      So, the result is: $\frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}$
    Now substitute $u$ back in:
    
    $\frac{\log{\left(u \right)}^{2}}{2}$
  Now substitute $u$ back in:
  
  $\frac{\log{\left(x + 1 \right)}^{2}}{2}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$${{\left(\log \left(x+1\right)\right)^2}\over{2}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

   2   
log (2)
-------
   2

$${{\left(\log 2\right)^2}\over{2}}$$

   2   
log (2)
-------
   2

$$\frac{\log{\left(2 \right)}^{2}}{2}$$

Simplify

Numerical answer [src]

0.240226506959101

0.240226506959101

The graph

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

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How to use it?

Identical expressions

Similar expressions

What you mean?

You entered:

What you mean?

Integral of log(x+1)/(x+1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2