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Identical expressions
(x^ two +x+ one)e^(3x)
(x squared plus x plus 1)e to the power of (3x)
(x to the power of two plus x plus one)e to the power of (3x)
(x2+x+1)e(3x)
x2+x+1e3x
(x²+x+1)e^(3x)
(x to the power of 2+x+1)e to the power of (3x)
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Similar expressions
(x^2-x+1)e^(3x)
(x^2+x-1)e^(3x)

Solving integrals/ (x^2+x+1)e^(3x)

Integral of (x^2+x+1)e^(3x) dx

The solution

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

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(x^{2} + x + 1\right) e^{3 x} = x^{2} e^{3 x} + x e^{3 x} + e^{3 x}$
2. Integrate term-by-term:
  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. There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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:
      
      $\frac{e^{3 x}}{3}$
      
      Method #2
      
      Let $u = e^{3 x}$ .
      
      Then let $du = 3 e^{3 x} dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{1}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{3}\, du = \frac{\int 1\, du}{3}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{3}$
      
      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 $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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 $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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: $\frac{2 e^{3 x}}{27}$
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = e^{3 x}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 1$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{e^{3 x}}{3}\, dx = \frac{\int e^{3 x}\, dx}{3}$
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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: $\frac{e^{3 x}}{9}$
  1. Let $u = 3 x$ .
    
    Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
    
    $\int \frac{e^{u}}{9}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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:
    
    $\frac{e^{3 x}}{3}$
  The result is: $\frac{x^{2} e^{3 x}}{3} + \frac{x e^{3 x}}{9} + \frac{8 e^{3 x}}{27}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = x^{2} + x + 1$ and let $\operatorname{dv}{\left(x \right)} = e^{3 x}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 2 x + 1$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 3 x$ .
    
    Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
    
    $\int \frac{e^{u}}{9}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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 $u{\left(x \right)} = \frac{2 x}{3} + \frac{1}{3}$ 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 $\frac{du}{3}$ :
    
    $\int \frac{e^{u}}{9}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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 $\frac{du}{3}$ :
    
    $\int \frac{e^{u}}{9}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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: $\frac{2 e^{3 x}}{27}$
Method #3
1. Rewrite the integrand:
  
  $\left(x^{2} + x + 1\right) e^{3 x} = x^{2} e^{3 x} + x e^{3 x} + e^{3 x}$
2. Integrate term-by-term:
  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 $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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 $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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 $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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: $\frac{2 e^{3 x}}{27}$
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = e^{3 x}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 1$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{e^{3 x}}{3}\, dx = \frac{\int e^{3 x}\, dx}{3}$
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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: $\frac{e^{3 x}}{9}$
  1. Let $u = 3 x$ .
    
    Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
    
    $\int \frac{e^{u}}{9}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      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:
    
    $\frac{e^{3 x}}{3}$
  The result is: $\frac{x^{2} e^{3 x}}{3} + \frac{x e^{3 x}}{9} + \frac{8 e^{3 x}}{27}$
Now simplify:

$\frac{\left(9 x^{2} + 3 x + 8\right) e^{3 x}}{27}$
Add the constant of integration:

$\frac{\left(9 x^{2} + 3 x + 8\right) e^{3 x}}{27}+ \mathrm{constant}$

The answer is:

$\frac{\left(9 x^{2} + 3 x + 8\right) e^{3 x}}{27}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                    
 |                               3*x    2  3*x      3*x
 | / 2        \  3*x          8*e      x *e      x*e   
 | \x  + x + 1/*e    dx = C + ------ + ------- + ------
 |                              27        3        9   
/

$${{\left(9\,x^2-6\,x+2\right)\,e^{3\,x}}\over{27}}+{{\left(3\,x-1 \right)\,e^{3\,x}}\over{9}}+{{e^{3\,x}}\over{3}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

           3
  8    20*e 
- -- + -----
  27     27

$${{20\,e^3}\over{27}}-{{8}\over{27}}$$

           3
  8    20*e 
- -- + -----
  27     27

$$- \frac{8}{27} + \frac{20 e^{3}}{27}$$

Simplify

Numerical answer [src]

14.5818792023612

14.5818792023612

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (x^2+x+1)e^(3x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3