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Identical expressions
(xe^(zero .3x)-ln(x+ two))dx
(xe to the power of (0.3x) minus ln(x plus 2))dx
(xe to the power of (zero .3x) minus ln(x plus two))dx
(xe(0.3x)-ln(x+2))dx
xe0.3x-lnx+2dx
xe^0.3x-lnx+2dx
Similar expressions
(xe^(0.3x)+ln(x+2))dx
(xe^(0.3x)-ln(x-2))dx

Solving integrals/ (xe^(0.3x)-ln(x+2))dx

Integral of (xe^(0.3x)-ln(x+2))dx dx

The solution

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 10                         
  /                         
 |                          
 |  /   3*x             \   
 |  |   ---             |   
 |  |    10             |   
 |  \x*E    - log(x + 2)/ dx
 |                          
/                           
0

$$\int\limits_{0}^{10} \left(e^{\frac{3 x}{10}} x - \log{\left(x + 2 \right)}\right)\, dx$$

Integral(x*E^(3*x/10) - log(x + 2), (x, 0, 10))

Detail solution

Integrate term-by-term:
1. Don't know the steps in finding this integral.
  
  But the integral is
  
  $\frac{\left(30 x - 100\right) e^{\frac{3 x}{10}}}{9}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \log{\left(x + 2 \right)}\right)\, dx = - \int \log{\left(x + 2 \right)}\, dx$
  1. There are multiple ways to do this integral.
    Method #1
    1. Let $u = x + 2$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \log{\left(u \right)}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = \log{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = 1$ .
      
      Then $\operatorname{du}{\left(u \right)} = \frac{1}{u}$ .
      
      To find $v{\left(u \right)}$ :
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      Now evaluate the sub-integral.
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      Now substitute $u$ back in:
      
      $- x + \left(x + 2\right) \log{\left(x + 2 \right)} - 2$
    Method #2
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = \log{\left(x + 2 \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{1}{x + 2}$ .
      
      To find $v{\left(x \right)}$ :
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
      
      Now evaluate the sub-integral.
    2. Rewrite the integrand:
      
      $\frac{x}{x + 2} = 1 - \frac{2}{x + 2}$
    3. Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{2}{x + 2}\right)\, dx = - 2 \int \frac{1}{x + 2}\, dx$
      
      Let $u = x + 2$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x + 2 \right)}$
      
      So, the result is: $- 2 \log{\left(x + 2 \right)}$
      
      The result is: $x - 2 \log{\left(x + 2 \right)}$
  So, the result is: $x - \left(x + 2\right) \log{\left(x + 2 \right)} + 2$
The result is: $x - \left(x + 2\right) \log{\left(x + 2 \right)} + \frac{\left(30 x - 100\right) e^{\frac{3 x}{10}}}{9} + 2$
Now simplify:

$x - \left(x + 2\right) \log{\left(x + 2 \right)} + \frac{10 \left(3 x - 10\right) e^{\frac{3 x}{10}}}{9} + 2$
Add the constant of integration:

$x - \left(x + 2\right) \log{\left(x + 2 \right)} + \frac{10 \left(3 x - 10\right) e^{\frac{3 x}{10}}}{9} + 2+ \mathrm{constant}$

The answer is:

$x - \left(x + 2\right) \log{\left(x + 2 \right)} + \frac{10 \left(3 x - 10\right) e^{\frac{3 x}{10}}}{9} + 2+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                              
 |                                                                            3*x
 | /   3*x             \                                                      ---
 | |   ---             |                                                       10
 | |    10             |                                       (-100 + 30*x)*e   
 | \x*E    - log(x + 2)/ dx = 2 + C + x - (x + 2)*log(x + 2) + ------------------
 |                                                                     9         
/

$$\int \left(e^{\frac{3 x}{10}} x - \log{\left(x + 2 \right)}\right)\, dx = C + x - \left(x + 2\right) \log{\left(x + 2 \right)} + \frac{\left(30 x - 100\right) e^{\frac{3 x}{10}}}{9} + 2$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                                   3
190                           200*e 
--- - 12*log(12) + 2*log(2) + ------
 9                              9

$$- 12 \log{\left(12 \right)} + 2 \log{\left(2 \right)} + \frac{190}{9} + \frac{200 e^{3}}{9}$$

                                   3
190                           200*e 
--- - 12*log(12) + 2*log(2) + ------
 9                              9

$$- 12 \log{\left(12 \right)} + 2 \log{\left(2 \right)} + \frac{190}{9} + \frac{200 e^{3}}{9}$$

190/9 - 12*log(12) + 2*log(2) + 200*exp(3)/9

Numerical answer [src]

439.023790634501

439.023790634501

The graph

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

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Identical expressions

Similar expressions

Integral of (xe^(0.3x)-ln(x+2))dx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2