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x*e^(x^ two - one)
x multiply by e to the power of (x squared minus 1)
x multiply by e to the power of (x to the power of two minus one)
x*e(x2-1)
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x*e^(x^2-1)dx
Similar expressions
x*e^(x^2+1)
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Integral of x*e^(x^2-1) dx

The solution

You have entered [src]

  1             
  /             
 |              
 |      2       
 |     x  - 1   
 |  x*E       dx
 |              
/               
0

$$\int\limits_{0}^{1} e^{x^{2} - 1} x\, dx$$

Integral(x*E^(x^2 - 1), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = x^{2} - 1$ .
  
  Then let $du = 2 x dx$ and substitute $\frac{du}{2}$ :
  
  $\int \frac{e^{u}}{2}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\text{False}$
    1. 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:
  
  $\frac{e^{x^{2} - 1}}{2}$
Method #2
1. Rewrite the integrand:
  
  $e^{x^{2} - 1} x = \frac{x e^{x^{2}}}{e}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{x e^{x^{2}}}{e}\, dx = \frac{\int x e^{x^{2}}\, dx}{e}$
  1. Let $u = x^{2}$ .
    
    Then let $du = 2 x dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{e^{u}}{2}\, du$
    1. 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:
    
    $\frac{e^{x^{2}}}{2}$
  So, the result is: $\frac{e^{x^{2}}}{2 e}$
Method #3
1. Rewrite the integrand:
  
  $e^{x^{2} - 1} x = \frac{x e^{x^{2}}}{e}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{x e^{x^{2}}}{e}\, dx = \frac{\int x e^{x^{2}}\, dx}{e}$
  1. Let $u = x^{2}$ .
    
    Then let $du = 2 x dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{e^{u}}{2}\, du$
    1. 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:
    
    $\frac{e^{x^{2}}}{2}$
  So, the result is: $\frac{e^{x^{2}}}{2 e}$
Now simplify:

$\frac{e^{x^{2} - 1}}{2}$
Add the constant of integration:

$\frac{e^{x^{2} - 1}}{2}+ \mathrm{constant}$

The answer is:

$\frac{e^{x^{2} - 1}}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                          
 |                      2    
 |     2               x  - 1
 |    x  - 1          e      
 | x*E       dx = C + -------
 |                       2   
/

$$\int e^{x^{2} - 1} x\, dx = C + \frac{e^{x^{2} - 1}}{2}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

     -1
1   e  
- - ---
2    2

$$\frac{1}{2} - \frac{1}{2 e}$$

     -1
1   e  
- - ---
2    2

$$\frac{1}{2} - \frac{1}{2 e}$$

1/2 - exp(-1)/2

Numerical answer [src]

0.316060279414279

0.316060279414279

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

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How to use it?

Identical expressions

Similar expressions

Integral of x*e^(x^2-1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3