Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of x*2^x
Integral of x*sin(x/2)
Integral of x^4/(x^2+1)
Integral of x^2*a^x
Identical expressions
(x^ two +3x- one)e^(2x)
(x squared plus 3x minus 1)e to the power of (2x)
(x to the power of two plus 3x minus one)e to the power of (2x)
(x2+3x-1)e(2x)
x2+3x-1e2x
(x²+3x-1)e^(2x)
(x to the power of 2+3x-1)e to the power of (2x)
x^2+3x-1e^2x
(x^2+3x-1)e^(2x)dx
Similar expressions
(x^2+3x+1)e^(2x)
(x^2-3x-1)e^(2x)

Solving integrals/ (x^2+3x-1)e^(2x)

Integral of (x^2+3x-1)e^(2x) dx

The solution

You have entered [src]

  1                       
  /                       
 |                        
 |  / 2          \  2*x   
 |  \x  + 3*x - 1/*e    dx
 |                        
/                         
0

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

Simplify

Detail solution

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

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                                    
 |                                               2  2*x
 | / 2          \  2*x           2*x      2*x   x *e   
 | \x  + 3*x - 1/*e    dx = C - e    + x*e    + -------
 |                                                 2   
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

$${{e^2}\over{2}}+1$$

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

Simplify

Numerical answer [src]

4.69452804946533

4.69452804946533

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^2+3x-1)e^(2x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3