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x^ two *exp(3x+ one)
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Similar expressions
x^2*exp(3x-1)

Integral of x^2*exp(3x+1) dx

The solution

You have entered [src]

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

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $x^{2} e^{3 x + 1} = e x^{2} e^{3 x}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int e x^{2} e^{3 x}\, dx = e \int x^{2} e^{3 x}\, dx$
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = e^{3 x}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 2 x$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 x}}{3}$
    Now evaluate the sub-integral.
  2. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = \frac{2 x}{3}$ and let $\operatorname{dv}{\left(x \right)} = e^{3 x}$ .
    
    Then $\operatorname{du}{\left(x \right)} = \frac{2}{3}$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 x}}{3}$
    Now evaluate the sub-integral.
  3. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{2 e^{3 x}}{9}\, dx = \frac{2 \int e^{3 x}\, dx}{9}$
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 x}}{3}$
    So, the result is: $\frac{2 e^{3 x}}{27}$
  So, the result is: $e \left(\frac{x^{2} e^{3 x}}{3} - \frac{2 x e^{3 x}}{9} + \frac{2 e^{3 x}}{27}\right)$
Method #2
1. Rewrite the integrand:
  
  $x^{2} e^{3 x + 1} = e x^{2} e^{3 x}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int e x^{2} e^{3 x}\, dx = e \int x^{2} e^{3 x}\, dx$
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = e^{3 x}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 2 x$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 x}}{3}$
    Now evaluate the sub-integral.
  2. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = \frac{2 x}{3}$ and let $\operatorname{dv}{\left(x \right)} = e^{3 x}$ .
    
    Then $\operatorname{du}{\left(x \right)} = \frac{2}{3}$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 x}}{3}$
    Now evaluate the sub-integral.
  3. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{2 e^{3 x}}{9}\, dx = \frac{2 \int e^{3 x}\, dx}{9}$
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{3}\, du$
      
      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: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 x}}{3}$
    So, the result is: $\frac{2 e^{3 x}}{27}$
  So, the result is: $e \left(\frac{x^{2} e^{3 x}}{3} - \frac{2 x e^{3 x}}{9} + \frac{2 e^{3 x}}{27}\right)$
Now simplify:

$\frac{\left(9 x^{2} - 6 x + 2\right) e^{3 x + 1}}{27}$
Add the constant of integration:

$\frac{\left(9 x^{2} - 6 x + 2\right) e^{3 x + 1}}{27}+ \mathrm{constant}$

The answer is:

$\frac{\left(9 x^{2} - 6 x + 2\right) e^{3 x + 1}}{27}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                    
 |                        /   3*x        3*x    2  3*x\
 |  2  3*x + 1            |2*e      2*x*e      x *e   |
 | x *e        dx = C + E*|------ - -------- + -------|
 |                        \  27        9          3   /
/

$$\int x^{2} e^{3 x + 1}\, dx = C + e \left(\frac{x^{2} e^{3 x}}{3} - \frac{2 x e^{3 x}}{9} + \frac{2 e^{3 x}}{27}\right)$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

           4
  2*E   5*e 
- --- + ----
   27    27

$$- \frac{2 e}{27} + \frac{5 e^{4}}{27}$$

           4
  2*E   5*e 
- --- + ----
   27    27

$$- \frac{2 e}{27} + \frac{5 e^{4}}{27}$$

-2*E/27 + 5*exp(4)/27

Numerical answer [src]

9.90941431514086

9.90941431514086

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^2*exp(3x+1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2