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Identical expressions
(6x+ two)*e^(two -9x)
(6x plus 2) multiply by e to the power of (2 minus 9x)
(6x plus two) multiply by e to the power of (two minus 9x)
(6x+2)*e(2-9x)
6x+2*e2-9x
(6x+2)e^(2-9x)
(6x+2)e(2-9x)
6x+2e2-9x
6x+2e^2-9x
(6x+2)*e^(2-9x)dx
Similar expressions
(6x-2)*e^(2-9x)
(6x+2)*e^(2+9x)

Integral of (6x+2)*e^(2-9x) dx

The solution

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  1                      
  /                      
 |                       
 |             2 - 9*x   
 |  (6*x + 2)*E        dx
 |                       
/                        
0

$$\int\limits_{0}^{1} e^{2 - 9 x} \left(6 x + 2\right)\, dx$$

Integral((6*x + 2)*E^(2 - 9*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $e^{2 - 9 x} \left(6 x + 2\right) = 6 x e^{2} e^{- 9 x} + 2 e^{2} e^{- 9 x}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 6 x e^{2} e^{- 9 x}\, dx = 6 e^{2} \int x e^{- 9 x}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = e^{- 9 x}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = - 9 x$ .
      
      Then let $du = - 9 dx$ and substitute $- \frac{du}{9}$ :
      
      $\int \left(- \frac{e^{u}}{9}\right)\, du$
      
      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: $- \frac{e^{u}}{9}$
      
      Now substitute $u$ back in:
      
      $- \frac{e^{- 9 x}}{9}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{- 9 x}}{9}\right)\, dx = - \frac{\int e^{- 9 x}\, dx}{9}$
      
      Let $u = - 9 x$ .
      
      Then let $du = - 9 dx$ and substitute $- \frac{du}{9}$ :
      
      $\int \left(- \frac{e^{u}}{9}\right)\, du$
      
      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: $- \frac{e^{u}}{9}$
      
      Now substitute $u$ back in:
      
      $- \frac{e^{- 9 x}}{9}$
      
      So, the result is: $\frac{e^{- 9 x}}{81}$
    So, the result is: $6 \left(- \frac{x e^{- 9 x}}{9} - \frac{e^{- 9 x}}{81}\right) e^{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 e^{2} e^{- 9 x}\, dx = 2 e^{2} \int e^{- 9 x}\, dx$
    1. Let $u = - 9 x$ .
      
      Then let $du = - 9 dx$ and substitute $- \frac{du}{9}$ :
      
      $\int \left(- \frac{e^{u}}{9}\right)\, du$
      
      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: $- \frac{e^{u}}{9}$
      
      Now substitute $u$ back in:
      
      $- \frac{e^{- 9 x}}{9}$
    So, the result is: $- \frac{2 e^{2} e^{- 9 x}}{9}$
  The result is: $6 \left(- \frac{x e^{- 9 x}}{9} - \frac{e^{- 9 x}}{81}\right) e^{2} - \frac{2 e^{2} e^{- 9 x}}{9}$
Method #2
1. Rewrite the integrand:
  
  $e^{2 - 9 x} \left(6 x + 2\right) = 6 x e^{2} e^{- 9 x} + 2 e^{2} e^{- 9 x}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 6 x e^{2} e^{- 9 x}\, dx = 6 e^{2} \int x e^{- 9 x}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = e^{- 9 x}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = - 9 x$ .
      
      Then let $du = - 9 dx$ and substitute $- \frac{du}{9}$ :
      
      $\int \left(- \frac{e^{u}}{9}\right)\, du$
      
      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: $- \frac{e^{u}}{9}$
      
      Now substitute $u$ back in:
      
      $- \frac{e^{- 9 x}}{9}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{- 9 x}}{9}\right)\, dx = - \frac{\int e^{- 9 x}\, dx}{9}$
      
      Let $u = - 9 x$ .
      
      Then let $du = - 9 dx$ and substitute $- \frac{du}{9}$ :
      
      $\int \left(- \frac{e^{u}}{9}\right)\, du$
      
      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: $- \frac{e^{u}}{9}$
      
      Now substitute $u$ back in:
      
      $- \frac{e^{- 9 x}}{9}$
      
      So, the result is: $\frac{e^{- 9 x}}{81}$
    So, the result is: $6 \left(- \frac{x e^{- 9 x}}{9} - \frac{e^{- 9 x}}{81}\right) e^{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 e^{2} e^{- 9 x}\, dx = 2 e^{2} \int e^{- 9 x}\, dx$
    1. Let $u = - 9 x$ .
      
      Then let $du = - 9 dx$ and substitute $- \frac{du}{9}$ :
      
      $\int \left(- \frac{e^{u}}{9}\right)\, du$
      
      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: $- \frac{e^{u}}{9}$
      
      Now substitute $u$ back in:
      
      $- \frac{e^{- 9 x}}{9}$
    So, the result is: $- \frac{2 e^{2} e^{- 9 x}}{9}$
  The result is: $6 \left(- \frac{x e^{- 9 x}}{9} - \frac{e^{- 9 x}}{81}\right) e^{2} - \frac{2 e^{2} e^{- 9 x}}{9}$
Now simplify:

$- \frac{2 \left(9 x + 4\right) e^{2 - 9 x}}{27}$
Add the constant of integration:

$- \frac{2 \left(9 x + 4\right) e^{2 - 9 x}}{27}+ \mathrm{constant}$

The answer is:

$- \frac{2 \left(9 x + 4\right) e^{2 - 9 x}}{27}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                 
 |                               /   -9*x      -9*x\         2  -9*x
 |            2 - 9*x            |  e       x*e    |  2   2*e *e    
 | (6*x + 2)*E        dx = C + 6*|- ----- - -------|*e  - ----------
 |                               \    81       9   /          9     
/

$$\int e^{2 - 9 x} \left(6 x + 2\right)\, dx = C + 6 \left(- \frac{x e^{- 9 x}}{9} - \frac{e^{- 9 x}}{81}\right) e^{2} - \frac{2 e^{2} e^{- 9 x}}{9}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

      -7      2
  26*e     8*e 
- ------ + ----
    27      27

$$- \frac{26}{27 e^{7}} + \frac{8 e^{2}}{27}$$

      -7      2
  26*e     8*e 
- ------ + ----
    27      27

$$- \frac{26}{27 e^{7}} + \frac{8 e^{2}}{27}$$

-26*exp(-7)/27 + 8*exp(2)/27

Numerical answer [src]

2.18847184667929

2.18847184667929

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

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Identical expressions

Similar expressions

Integral of (6x+2)*e^(2-9x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2