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Integral of d{x}:
Integral of e^x/x^2
Integral of 1/√(4-x^2)
Integral of a^x*e^x
Integral of 1/(x^2+2*x+5)
Identical expressions
(two -3x)e^(4x- five)
(2 minus 3x)e to the power of (4x minus 5)
(two minus 3x)e to the power of (4x minus five)
(2-3x)e(4x-5)
2-3xe4x-5
2-3xe^4x-5
(2-3x)e^(4x-5)dx
Similar expressions
(2-3x)e^(4x+5)
(2+3x)e^(4x-5)

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

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

The solution

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 5/8                     
  /                      
 |                       
 |             4*x - 5   
 |  (2 - 3*x)*e        dx
 |                       
/                        
-oo

$$\int\limits_{-\infty}^{\frac{5}{8}} \left(- 3 x + 2\right) e^{4 x - 5}\, dx$$

Integral((2 - 3*x)*E^(4*x - 1*5), (x, -oo, 5/8))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(2 - 3 x\right) e^{4 x - 5} = - \frac{3 x e^{4 x}}{e^{5}} + \frac{2 e^{4 x}}{e^{5}}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{3 x e^{4 x}}{e^{5}}\right)\, dx = - \frac{3 \int x e^{4 x}\, dx}{e^{5}}$
    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 $\operatorname{dv}{\left(x \right)} = e^{4 x}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{e^{u}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{4 x}}{4}$
      
      Method #2
      
      Let $u = e^{4 x}$ .
      
      Then let $du = 4 e^{4 x} dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{1}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{4}\, du = \frac{\int 1\, du}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{4 x}}{4}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{4 x}}{4}\, dx = \frac{\int e^{4 x}\, dx}{4}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{e^{u}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{4 x}}{4}$
      
      So, the result is: $\frac{e^{4 x}}{16}$
    So, the result is: $- \frac{3 \left(\frac{x e^{4 x}}{4} - \frac{e^{4 x}}{16}\right)}{e^{5}}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{2 e^{4 x}}{e^{5}}\, dx = \frac{2 \int e^{4 x}\, dx}{e^{5}}$
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{e^{u}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{4 x}}{4}$
    So, the result is: $\frac{e^{4 x}}{2 e^{5}}$
  The result is: $- \frac{3 \left(\frac{x e^{4 x}}{4} - \frac{e^{4 x}}{16}\right)}{e^{5}} + \frac{e^{4 x}}{2 e^{5}}$
Method #2
1. Rewrite the integrand:
  
  $\left(2 - 3 x\right) e^{4 x - 5} = - \frac{3 x e^{4 x}}{e^{5}} + \frac{2 e^{4 x}}{e^{5}}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{3 x e^{4 x}}{e^{5}}\right)\, dx = - \frac{3 \int x e^{4 x}\, dx}{e^{5}}$
    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 $\operatorname{dv}{\left(x \right)} = e^{4 x}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{e^{u}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{4 x}}{4}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{4 x}}{4}\, dx = \frac{\int e^{4 x}\, dx}{4}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{e^{u}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{4 x}}{4}$
      
      So, the result is: $\frac{e^{4 x}}{16}$
    So, the result is: $- \frac{3 \left(\frac{x e^{4 x}}{4} - \frac{e^{4 x}}{16}\right)}{e^{5}}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{2 e^{4 x}}{e^{5}}\, dx = \frac{2 \int e^{4 x}\, dx}{e^{5}}$
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{e^{u}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{4 x}}{4}$
    So, the result is: $\frac{e^{4 x}}{2 e^{5}}$
  The result is: $- \frac{3 \left(\frac{x e^{4 x}}{4} - \frac{e^{4 x}}{16}\right)}{e^{5}} + \frac{e^{4 x}}{2 e^{5}}$
Now simplify:

$\frac{\left(11 - 12 x\right) e^{4 x - 5}}{16}$
Add the constant of integration:

$\frac{\left(11 - 12 x\right) e^{4 x - 5}}{16}+ \mathrm{constant}$

The answer is:

$\frac{\left(11 - 12 x\right) e^{4 x - 5}}{16}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                              
 |                              -5  4*x     /   4*x      4*x\    
 |            4*x - 5          e  *e        |  e      x*e   |  -5
 | (2 - 3*x)*e        dx = C + -------- - 3*|- ---- + ------|*e  
 |                                2         \   16      4   /    
/

$${{e^{4\,x-5}}\over{2}}-{{3\,\left(4\,x-1\right)\,e^{4\,x-5}}\over{ 16}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

   -5/2
7*e    
-------
   32

$${{7\,e^ {- {{5}\over{2}} }}\over{32}}-{{\left(12\,{\it oo}+11 \right)\,e^{-4\,{\it oo}-5}}\over{16}}$$

   -5/2
7*e    
-------
   32

$$\frac{7}{32 e^{\frac{5}{2}}}$$

Simplify

The graph

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2