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Identical expressions
xexp(4x^ two + three)
x exponent of (4x squared plus 3)
x exponent of (4x to the power of two plus three)
xexp(4x2+3)
xexp4x2+3
xexp(4x²+3)
xexp(4x to the power of 2+3)
xexp4x^2+3
xexp(4x^2+3)dx
Similar expressions
xexp(4x^2-3)

Solving integrals/ xexp(4x^2+3)

Integral of xexp(4x^2+3) dx

The solution

You have entered [src]

  1               
  /               
 |                
 |        2       
 |     4*x  + 3   
 |  x*e         dx
 |                
/                 
0

$$\int\limits_{0}^{1} x e^{4 x^{2} + 3}\, dx$$

Integral(x*exp(4*x^2 + 3), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = 4 x^{2} + 3$ .
  
  Then let $du = 8 x dx$ and substitute $\frac{du}{8}$ :
  
  $\int \frac{e^{u}}{8}\, du$
  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: $\frac{e^{u}}{8}$
  Now substitute $u$ back in:
  
  $\frac{e^{4 x^{2} + 3}}{8}$
Method #2
1. Rewrite the integrand:
  
  $x e^{4 x^{2} + 3} = x e^{3} e^{4 x^{2}}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int x e^{3} e^{4 x^{2}}\, dx = e^{3} \int x e^{4 x^{2}}\, dx$
  1. Let $u = 4 x^{2}$ .
    
    Then let $du = 8 x dx$ and substitute $\frac{du}{8}$ :
    
    $\int \frac{e^{u}}{8}\, du$
    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: $\frac{e^{u}}{8}$
    Now substitute $u$ back in:
    
    $\frac{e^{4 x^{2}}}{8}$
  So, the result is: $\frac{e^{3} e^{4 x^{2}}}{8}$
Method #3
1. Rewrite the integrand:
  
  $x e^{4 x^{2} + 3} = x e^{3} e^{4 x^{2}}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int x e^{3} e^{4 x^{2}}\, dx = e^{3} \int x e^{4 x^{2}}\, dx$
  1. Let $u = 4 x^{2}$ .
    
    Then let $du = 8 x dx$ and substitute $\frac{du}{8}$ :
    
    $\int \frac{e^{u}}{8}\, du$
    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: $\frac{e^{u}}{8}$
    Now substitute $u$ back in:
    
    $\frac{e^{4 x^{2}}}{8}$
  So, the result is: $\frac{e^{3} e^{4 x^{2}}}{8}$
Now simplify:

$\frac{e^{4 x^{2} + 3}}{8}$
Add the constant of integration:

$\frac{e^{4 x^{2} + 3}}{8}+ \mathrm{constant}$

The answer is:

$\frac{e^{4 x^{2} + 3}}{8}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                              
 |                          2    
 |       2               4*x  + 3
 |    4*x  + 3          e        
 | x*e         dx = C + ---------
 |                          8    
/

$$\int x e^{4 x^{2} + 3}\, dx = C + \frac{e^{4 x^{2} + 3}}{8}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

   3    7
  e    e 
- -- + --
  8    8

$$- \frac{e^{3}}{8} + \frac{e^{7}}{8}$$

   3    7
  e    e 
- -- + --
  8    8

$$- \frac{e^{3}}{8} + \frac{e^{7}}{8}$$

-exp(3)/8 + exp(7)/8

Numerical answer [src]

134.568452688159

134.568452688159

The graph

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

Other calculators

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Identical expressions

Similar expressions

Integral of xexp(4x^2+3) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3