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Identical expressions
(4x+ five)*e^(-3x)
(4x plus 5) multiply by e to the power of ( minus 3x)
(4x plus five) multiply by e to the power of ( minus 3x)
(4x+5)*e(-3x)
4x+5*e-3x
(4x+5)e^(-3x)
(4x+5)e(-3x)
4x+5e-3x
4x+5e^-3x
(4x+5)*e^(-3x)dx
Similar expressions
(4x-5)*e^(-3x)
(4x+5)*e^(3x)

Solving integrals/ (4x+5)*e^(-3x)

Integral of (4x+5)*e^(-3x) dx

The solution

You have entered [src]

  1                   
  /                   
 |                    
 |             -3*x   
 |  (4*x + 5)*e     dx
 |                    
/                     
0

$$\int\limits_{0}^{1} \left(4 x + 5\right) e^{- 3 x}\, dx$$

Integral((4*x + 5)/E^(3*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = 4 x + 5$ and let $\operatorname{dv}{\left(x \right)} = e^{- 3 x}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 4$ .
  
  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:
    
    $\int \left(- \frac{e^{u}}{3}\right)\, du = - \frac{\int e^{u}\, du}{3}$
    
    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:
    
    $\int \left(- \frac{1}{3}\right)\, du = - \frac{\int 1\, du}{3}$
    
    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. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{4 e^{- 3 x}}{3}\right)\, dx = - \frac{4 \int e^{- 3 x}\, dx}{3}$
  1. Let $u = - 3 x$ .
    
    Then let $du = - 3 dx$ and substitute $- \frac{du}{3}$ :
    
    $\int \frac{e^{u}}{9}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{u}}{3}\right)\, du = - \frac{\int e^{u}\, du}{3}$
      
      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: $\frac{4 e^{- 3 x}}{9}$
Method #2
1. Rewrite the integrand:
  
  $\left(4 x + 5\right) e^{- 3 x} = 4 x e^{- 3 x} + 5 e^{- 3 x}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 4 x e^{- 3 x}\, dx = 4 \int x e^{- 3 x}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = e^{- 3 x}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      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:
      
      $\int \left(- \frac{e^{u}}{3}\right)\, du = - \frac{\int e^{u}\, du}{3}$
      
      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:
      
      $\int \left(- \frac{e^{- 3 x}}{3}\right)\, dx = - \frac{\int e^{- 3 x}\, dx}{3}$
      
      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:
      
      $\int \left(- \frac{e^{u}}{3}\right)\, du = - \frac{\int e^{u}\, du}{3}$
      
      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: $\frac{e^{- 3 x}}{9}$
    So, the result is: $- \frac{4 x e^{- 3 x}}{3} - \frac{4 e^{- 3 x}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 5 e^{- 3 x}\, dx = 5 \int e^{- 3 x}\, dx$
    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:
      
      $\int \left(- \frac{e^{u}}{3}\right)\, du = - \frac{\int e^{u}\, du}{3}$
      
      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: $- \frac{5 e^{- 3 x}}{3}$
  The result is: $- \frac{4 x e^{- 3 x}}{3} - \frac{19 e^{- 3 x}}{9}$
Now simplify:

$\frac{\left(- 12 x - 19\right) e^{- 3 x}}{9}$
Add the constant of integration:

$\frac{\left(- 12 x - 19\right) e^{- 3 x}}{9}+ \mathrm{constant}$

The answer is:

$\frac{\left(- 12 x - 19\right) e^{- 3 x}}{9}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                  
 |                             -3*x              -3*x
 |            -3*x          4*e       (5 + 4*x)*e    
 | (4*x + 5)*e     dx = C - ------- - ---------------
 |                             9             3       
/

$$-{{4\,\left(3\,x+1\right)\,e^ {- 3\,x }}\over{9}}-{{5\,e^ {- 3\,x } }\over{3}}$$

Simplify

The answer [src]

         -3
19   31*e  
-- - ------
9      9

$${{19}\over{9}}-{{31\,e^ {- 3 }}\over{9}}$$

         -3
19   31*e  
-- - ------
9      9

$$- \frac{31}{9 e^{3}} + \frac{19}{9}$$

Simplify

Numerical answer [src]

1.93962232006625

1.93962232006625

The graph

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

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Identical expressions

Similar expressions

Integral of (4x+5)*e^(-3x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2