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Integral of d{x}:
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Integral of (x+1)e^(2x)
Identical expressions
(x+ one)e^(2x)
(x plus 1)e to the power of (2x)
(x plus one)e to the power of (2x)
(x+1)e(2x)
x+1e2x
x+1e^2x
(x+1)e^(2x)dx
Similar expressions
(x-1)e^(2x)
(x^2-2*x+1)*e^(2*x)

Solving integrals/ (x+1)e^(2x)

Integral of (x+1)e^(2x) dx

The solution

You have entered [src]

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

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $e^{2 x} \left(x + 1\right) = x e^{2 x} + e^{2 x}$
2. Integrate term-by-term:
  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 $v{\left(x \right)}$ :
    1. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      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: $\frac{e^{u}}{2}$
      
      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:
    
    $\int \frac{e^{2 x}}{2}\, dx = \frac{\int e^{2 x}\, dx}{2}$
    1. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      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: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 x}}{2}$
    So, the result is: $\frac{e^{2 x}}{4}$
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{e^{u}}{2}\, du$
    1. 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: $\frac{e^{u}}{2}$
    Now substitute $u$ back in:
    
    $\frac{e^{2 x}}{2}$
  The result is: $\frac{x e^{2 x}}{2} + \frac{e^{2 x}}{4}$
Method #2
1. Rewrite the integrand:
  
  $e^{2 x} \left(x + 1\right) = x e^{2 x} + e^{2 x}$
2. Integrate term-by-term:
  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 $v{\left(x \right)}$ :
    1. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      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: $\frac{e^{u}}{2}$
      
      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:
    
    $\int \frac{e^{2 x}}{2}\, dx = \frac{\int e^{2 x}\, dx}{2}$
    1. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      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: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 x}}{2}$
    So, the result is: $\frac{e^{2 x}}{4}$
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{e^{u}}{2}\, du$
    1. 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: $\frac{e^{u}}{2}$
    Now substitute $u$ back in:
    
    $\frac{e^{2 x}}{2}$
  The result is: $\frac{x e^{2 x}}{2} + \frac{e^{2 x}}{4}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$$\int e^{2 x} \left(x + 1\right)\, dx = C + \frac{x e^{2 x}}{2} + \frac{e^{2 x}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

         2
  1   3*e 
- - + ----
  4    4

$$- \frac{1}{4} + \frac{3 e^{2}}{4}$$

         2
  1   3*e 
- - + ----
  4    4

$$- \frac{1}{4} + \frac{3 e^{2}}{4}$$

-1/4 + 3*exp(2)/4

Numerical answer [src]

5.29179207419799

5.29179207419799

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 (x+1)e^(2x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2