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Integral of d{x}:
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Identical expressions
(5x+ four)*e^(3x- two)
(5x plus 4) multiply by e to the power of (3x minus 2)
(5x plus four) multiply by e to the power of (3x minus two)
(5x+4)*e(3x-2)
5x+4*e3x-2
(5x+4)e^(3x-2)
(5x+4)e(3x-2)
5x+4e3x-2
5x+4e^3x-2
(5x+4)*e^(3x-2)dx
Similar expressions
(5x+4)*e^(3x+2)
(5x-4)*e^(3x-2)

Solving integrals/ (5x+4)*e^(3x-2)

Integral of (5x+4)*e^(3x-2) dx

The solution

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 2/3                     
  /                      
 |                       
 |             3*x - 2   
 |  (5*x + 4)*e        dx
 |                       
/                        
-oo

$$\int\limits_{-\infty}^{\frac{2}{3}} \left(5 x + 4\right) e^{3 x - 2}\, dx$$

Integral((5*x + 4)*E^(3*x - 1*2), (x, -oo, 2/3))

Detail solution

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

$\frac{\left(15 x + 7\right) e^{3 x - 2}}{9}$
Add the constant of integration:

$\frac{\left(15 x + 7\right) e^{3 x - 2}}{9}+ \mathrm{constant}$

The answer is:

$\frac{\left(15 x + 7\right) e^{3 x - 2}}{9}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                
 |                               /   3*x      3*x\          -2  3*x
 |            3*x - 2            |  e      x*e   |  -2   4*e  *e   
 | (5*x + 4)*e        dx = C + 5*|- ---- + ------|*e   + ----------
 |                               \   9       3   /           3     
/

$${{5\,\left(3\,x-1\right)\,e^{3\,x-2}}\over{9}}+{{4\,e^{3\,x-2} }\over{3}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

17/9

$${{\left(15\,{\it oo}-7\right)\,e^{-3\,{\it oo}-2}}\over{9}}+{{17 }\over{9}}$$

17/9

$$\frac{17}{9}$$

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

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Similar expressions

Integral of (5x+4)*e^(3x-2) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2