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Identical expressions
(four *x+ three)*e^(two *x)*dx
(4 multiply by x plus 3) multiply by e to the power of (2 multiply by x) multiply by dx
(four multiply by x plus three) multiply by e to the power of (two multiply by x) multiply by dx
(4*x+3)*e(2*x)*dx
4*x+3*e2*x*dx
(4x+3)e^(2x)dx
(4x+3)e(2x)dx
4x+3e2xdx
4x+3e^2xdx
Similar expressions
(4*x-3)*e^(2*x)*dx

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

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

The solution

You have entered [src]

  1                    
  /                    
 |                     
 |             2*x     
 |  (4*x + 3)*e   *1 dx
 |                     
/                      
0

$$\int\limits_{0}^{1} \left(4 x + 3\right) e^{2 x} 1\, dx$$

Simplify

Detail solution

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

$\frac{\left(4 x + 1\right) e^{2 x}}{2}$
Add the constant of integration:

$\frac{\left(4 x + 1\right) e^{2 x}}{2}+ \mathrm{constant}$

The answer is:

$\frac{\left(4 x + 1\right) e^{2 x}}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                         
 |                            2*x           
 |            2*x            e           2*x
 | (4*x + 3)*e   *1 dx = C + ---- + 2*x*e   
 |                            2             
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

         2
  1   5*e 
- - + ----
  2    2

$${{5\,e^2}\over{2}}-{{1}\over{2}}$$

         2
  1   5*e 
- - + ----
  2    2

$$- \frac{1}{2} + \frac{5 e^{2}}{2}$$

Simplify

Numerical answer [src]

17.9726402473266

17.9726402473266

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 (4x+3)e^(2x)dx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3

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