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Identical expressions
(e^-(t)+t- one)/t^ two
(e to the power of minus (t) plus t minus 1) divide by t squared
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(e-(t)+t-1)/t2
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Similar expressions
(e^-(t)+t+1)/t^2
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Integral of (e^-(t)+t-1)/t^2 dx

The solution

You have entered [src]

  1               
  /               
 |                
 |   -t           
 |  E   + t - 1   
 |  ----------- dt
 |        2       
 |       t        
 |                
/                 
0

$$\int\limits_{0}^{1} \frac{\left(t + e^{- t}\right) - 1}{t^{2}}\, dt$$

Integral((E^(-t) + t - 1)/t^2, (t, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = - t$ .
  
  Then let $du = - dt$ and substitute $du$ :
  
  $\int \frac{u - e^{u} + 1}{u^{2}}\, du$
  1. Rewrite the integrand:
    
    $\frac{u - e^{u} + 1}{u^{2}} = \frac{1}{u} - \frac{e^{u}}{u^{2}} + \frac{1}{u^{2}}$
  2. Integrate term-by-term:
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{u}}{u^{2}}\right)\, du = - \int \frac{e^{u}}{u^{2}}\, du$
      
      UpperGammaRule(a=1, e=-2, context=exp(_u)/_u**2, symbol=_u)
      
      So, the result is: $\frac{\operatorname{E}_{2}\left(- u\right)}{u}$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$
    The result is: $\log{\left(u \right)} + \frac{\operatorname{E}_{2}\left(- u\right)}{u} - \frac{1}{u}$
  Now substitute $u$ back in:
  
  $\log{\left(- t \right)} - \frac{\operatorname{E}_{2}\left(t\right)}{t} + \frac{1}{t}$
Method #2
1. Rewrite the integrand:
  
  $\frac{\left(t + e^{- t}\right) - 1}{t^{2}} = \frac{\left(t e^{t} - e^{t} + 1\right) e^{- t}}{t^{2}}$
2. Let $u = - t$ .
  
  Then let $du = - dt$ and substitute $du$ :
  
  $\int \frac{u - e^{u} + 1}{u^{2}}\, du$
  1. Rewrite the integrand:
    
    $\frac{u - e^{u} + 1}{u^{2}} = \frac{1}{u} - \frac{e^{u}}{u^{2}} + \frac{1}{u^{2}}$
  2. Integrate term-by-term:
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{u}}{u^{2}}\right)\, du = - \int \frac{e^{u}}{u^{2}}\, du$
      
      UpperGammaRule(a=1, e=-2, context=exp(_u)/_u**2, symbol=_u)
      
      So, the result is: $\frac{\operatorname{E}_{2}\left(- u\right)}{u}$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$
    The result is: $\log{\left(u \right)} + \frac{\operatorname{E}_{2}\left(- u\right)}{u} - \frac{1}{u}$
  Now substitute $u$ back in:
  
  $\log{\left(- t \right)} - \frac{\operatorname{E}_{2}\left(t\right)}{t} + \frac{1}{t}$
Method #3
1. Rewrite the integrand:
  
  $\frac{\left(t + e^{- t}\right) - 1}{t^{2}} = \frac{1}{t} - \frac{1}{t^{2}} + \frac{e^{- t}}{t^{2}}$
2. Integrate term-by-term:
  1. The integral of $\frac{1}{t}$ is $\log{\left(t \right)}$ .
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{t^{2}}\right)\, dt = - \int \frac{1}{t^{2}}\, dt$
    1. The integral of $t^{n}$ is $\frac{t^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{t^{2}}\, dt = - \frac{1}{t}$
    So, the result is: $\frac{1}{t}$
  The result is: $\log{\left(t \right)} - \frac{\operatorname{E}_{2}\left(t\right)}{t} + \frac{1}{t}$
Now simplify:

$\frac{t \log{\left(- t \right)} - \operatorname{E}_{2}\left(t\right) + 1}{t}$
Add the constant of integration:

$\frac{t \log{\left(- t \right)} - \operatorname{E}_{2}\left(t\right) + 1}{t}+ \mathrm{constant}$

The answer is:

$\frac{t \log{\left(- t \right)} - \operatorname{E}_{2}\left(t\right) + 1}{t}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                               
 |                                                
 |  -t                                            
 | E   + t - 1          1   expint(2, t)          
 | ----------- dt = C + - - ------------ + log(-t)
 |       2              t        t                
 |      t                                         
 |                                                
/

$$\int \frac{\left(t + e^{- t}\right) - 1}{t^{2}}\, dt = C + \log{\left(- t \right)} - \frac{\operatorname{E}_{2}\left(t\right)}{t} + \frac{1}{t}$$

Simplify

The answer [src]

-expint(2, 1) - pi*I + EulerGamma

$$- \operatorname{E}_{2}\left(1\right) + \gamma - i \pi$$

-expint(2, 1) - pi*I + EulerGamma

$$- \operatorname{E}_{2}\left(1\right) + \gamma - i \pi$$

-expint(2, 1) - pi*i + EulerGamma

Numerical answer [src]

0.428720158115632

0.428720158115632

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

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Identical expressions

Similar expressions

Integral of (e^-(t)+t-1)/t^2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3