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Similar expressions
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Integral of (2x-1)/(3^x) dx

The solution

You have entered [src]

 oo           
  /           
 |            
 |  2*x - 1   
 |  ------- dx
 |      x     
 |     3      
 |            
/             
1

$$\int\limits_{1}^{\infty} \frac{2 x - 1}{3^{x}}\, dx$$

Integral((2*x - 1)/3^x, (x, 1, oo))

Detail solution

Rewrite the integrand:

$\frac{2 x - 1}{3^{x}} = 2 \cdot 3^{- x} x - 3^{- x}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 2 \cdot 3^{- x} x\, dx = 2 \int 3^{- x} x\, dx$
  1. Don't know the steps in finding this integral.
    
    But the integral is
    
    $\frac{3^{- x} \left(- x \log{\left(3 \right)} - 1\right)}{\log{\left(3 \right)}^{2}}$
  So, the result is: $\frac{2 \cdot 3^{- x} \left(- x \log{\left(3 \right)} - 1\right)}{\log{\left(3 \right)}^{2}}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 3^{- x}\right)\, dx = - \int 3^{- x}\, dx$
  1. Let $u = - x$ .
    
    Then let $du = - dx$ and substitute $- du$ :
    
    $\int \left(- 3^{u}\right)\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3^{u}\, du = - \int 3^{u}\, du$
      1. The integral of an exponential function is itself divided by the natural logarithm of the base.
        
        $\int 3^{u}\, du = \frac{3^{u}}{\log{\left(3 \right)}}$
      So, the result is: $- \frac{3^{u}}{\log{\left(3 \right)}}$
    Now substitute $u$ back in:
    
    $- \frac{3^{- x}}{\log{\left(3 \right)}}$
  So, the result is: $\frac{3^{- x}}{\log{\left(3 \right)}}$
The result is: $\frac{2 \cdot 3^{- x} \left(- x \log{\left(3 \right)} - 1\right)}{\log{\left(3 \right)}^{2}} + \frac{3^{- x}}{\log{\left(3 \right)}}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$$\int \frac{2 x - 1}{3^{x}}\, dx = C + \frac{2 \cdot 3^{- x} \left(- x \log{\left(3 \right)} - 1\right)}{\log{\left(3 \right)}^{2}} + \frac{3^{- x}}{\log{\left(3 \right)}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-(-2 - log(3)) 
---------------
        2      
   3*log (3)

$$- \frac{-2 - \log{\left(3 \right)}}{3 \log{\left(3 \right)}^{2}}$$

-(-2 - log(3)) 
---------------
        2      
   3*log (3)

$$- \frac{-2 - \log{\left(3 \right)}}{3 \log{\left(3 \right)}^{2}}$$

-(-2 - log(3))/(3*log(3)^2)

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

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Integral of (2x-1)/(3^x) dx

Limits of integration:

The graph:

Piecewise:

The solution