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Similar expressions
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Integral of 1/(e^(-ln(t))+1) dx

The solution

You have entered [src]

  0                
  /                
 |                 
 |       1         
 |  ------------ dt
 |   -log(t)       
 |  E        + 1   
 |                 
/                  
0

$$\int\limits_{0}^{0} \frac{1}{1 + e^{- \log{\left(t \right)}}}\, dt$$

Integral(1/(E^(-log(t)) + 1), (t, 0, 0))

Detail solution

Rewrite the integrand:

$\frac{1}{1 + e^{- \log{\left(t \right)}}} = \frac{t}{t + 1}$
Rewrite the integrand:

$\frac{t}{t + 1} = 1 - \frac{1}{t + 1}$
Integrate term-by-term:
1. The integral of a constant is the constant times the variable of integration:
  
  $\int 1\, dt = t$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{1}{t + 1}\right)\, dt = - \int \frac{1}{t + 1}\, dt$
  1. Let $u = t + 1$ .
    
    Then let $du = dt$ and substitute $du$ :
    
    $\int \frac{1}{u}\, du$
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    Now substitute $u$ back in:
    
    $\log{\left(t + 1 \right)}$
  So, the result is: $- \log{\left(t + 1 \right)}$
The result is: $t - \log{\left(t + 1 \right)}$
Add the constant of integration:

$t - \log{\left(t + 1 \right)}+ \mathrm{constant}$

The answer is:

$t - \log{\left(t + 1 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                    
 |                                     
 |      1                              
 | ------------ dt = C + t - log(1 + t)
 |  -log(t)                            
 | E        + 1                        
 |                                     
/

$$\int \frac{1}{1 + e^{- \log{\left(t \right)}}}\, dt = C + t - \log{\left(t + 1 \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

$$0$$

Numerical answer [src]

0.0

0.0

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

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Identical expressions

Similar expressions

Integral of 1/(e^(-ln(t))+1) dx

Limits of integration:

The graph:

Piecewise:

The solution