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Integral of (2-x-2*y)^((-x)/2) dy

The solution

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  1                    
  /                    
 |                     
 |               -x    
 |               ---   
 |                2    
 |  (2 - x - 2*y)    dy
 |                     
/                      
0

$$\int\limits_{0}^{1} \left(- 2 y + \left(2 - x\right)\right)^{\frac{\left(-1\right) x}{2}}\, dy$$

Integral((2 - x - 2*y)^((-x)/2), (y, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = - 2 y + \left(2 - x\right)$ .
  
  Then let $du = - 2 dy$ and substitute $- \frac{du}{2}$ :
  
  $\int \left(- \frac{u^{- \frac{x}{2}}}{2}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int u^{- \frac{x}{2}}\, du = - \frac{\int u^{- \frac{x}{2}}\, du}{2}$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{- \frac{x}{2}}\, du = \begin{cases} \frac{u^{1 - \frac{x}{2}}}{1 - \frac{x}{2}} & \text{for}\: \frac{x}{2} \neq 1 \\\log{\left(u \right)} & \text{otherwise} \end{cases}$
    So, the result is: $- \frac{\begin{cases} \frac{u^{1 - \frac{x}{2}}}{1 - \frac{x}{2}} & \text{for}\: \frac{x}{2} \neq 1 \\\log{\left(u \right)} & \text{otherwise} \end{cases}}{2}$
  Now substitute $u$ back in:
  
  $- \frac{\begin{cases} \frac{\left(- 2 y + \left(2 - x\right)\right)^{1 - \frac{x}{2}}}{1 - \frac{x}{2}} & \text{for}\: \frac{x}{2} \neq 1 \\\log{\left(- 2 y + \left(2 - x\right) \right)} & \text{otherwise} \end{cases}}{2}$
Method #2
1. Rewrite the integrand:
  
  $\left(- 2 y + \left(2 - x\right)\right)^{\frac{\left(-1\right) x}{2}} = \left(- x - 2 y + 2\right)^{- \frac{x}{2}}$
2. Let $u = - x - 2 y + 2$ .
  
  Then let $du = - 2 dy$ and substitute $- \frac{du}{2}$ :
  
  $\int \left(- \frac{u^{- \frac{x}{2}}}{2}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int u^{- \frac{x}{2}}\, du = - \frac{\int u^{- \frac{x}{2}}\, du}{2}$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{- \frac{x}{2}}\, du = \begin{cases} \frac{u^{1 - \frac{x}{2}}}{1 - \frac{x}{2}} & \text{for}\: \frac{x}{2} \neq 1 \\\log{\left(u \right)} & \text{otherwise} \end{cases}$
    So, the result is: $- \frac{\begin{cases} \frac{u^{1 - \frac{x}{2}}}{1 - \frac{x}{2}} & \text{for}\: \frac{x}{2} \neq 1 \\\log{\left(u \right)} & \text{otherwise} \end{cases}}{2}$
  Now substitute $u$ back in:
  
  $- \frac{\begin{cases} \frac{\left(- x - 2 y + 2\right)^{1 - \frac{x}{2}}}{1 - \frac{x}{2}} & \text{for}\: \frac{x}{2} \neq 1 \\\log{\left(- x - 2 y + 2 \right)} & \text{otherwise} \end{cases}}{2}$
Method #3
1. Rewrite the integrand:
  
  $\left(- 2 y + \left(2 - x\right)\right)^{\frac{\left(-1\right) x}{2}} = \left(- 2 y + \left(2 - x\right)\right)^{- \frac{x}{2}}$
2. Let $u = - 2 y + \left(2 - x\right)$ .
  
  Then let $du = - 2 dy$ and substitute $- \frac{du}{2}$ :
  
  $\int \left(- \frac{u^{- \frac{x}{2}}}{2}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int u^{- \frac{x}{2}}\, du = - \frac{\int u^{- \frac{x}{2}}\, du}{2}$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{- \frac{x}{2}}\, du = \begin{cases} \frac{u^{1 - \frac{x}{2}}}{1 - \frac{x}{2}} & \text{for}\: \frac{x}{2} \neq 1 \\\log{\left(u \right)} & \text{otherwise} \end{cases}$
    So, the result is: $- \frac{\begin{cases} \frac{u^{1 - \frac{x}{2}}}{1 - \frac{x}{2}} & \text{for}\: \frac{x}{2} \neq 1 \\\log{\left(u \right)} & \text{otherwise} \end{cases}}{2}$
  Now substitute $u$ back in:
  
  $- \frac{\begin{cases} \frac{\left(- 2 y + \left(2 - x\right)\right)^{1 - \frac{x}{2}}}{1 - \frac{x}{2}} & \text{for}\: \frac{x}{2} \neq 1 \\\log{\left(- 2 y + \left(2 - x\right) \right)} & \text{otherwise} \end{cases}}{2}$
Now simplify:

$\begin{cases} \frac{\left(- x - 2 y + 2\right)^{1 - \frac{x}{2}}}{x - 2} & \text{for}\: x \neq 2 \\- \frac{\log{\left(- x - 2 y + 2 \right)}}{2} & \text{otherwise} \end{cases}$
Add the constant of integration:

$\begin{cases} \frac{\left(- x - 2 y + 2\right)^{1 - \frac{x}{2}}}{x - 2} & \text{for}\: x \neq 2 \\- \frac{\log{\left(- x - 2 y + 2 \right)}}{2} & \text{otherwise} \end{cases}+ \mathrm{constant}$

The answer is:

$\begin{cases} \frac{\left(- x - 2 y + 2\right)^{1 - \frac{x}{2}}}{x - 2} & \text{for}\: x \neq 2 \\- \frac{\log{\left(- x - 2 y + 2 \right)}}{2} & \text{otherwise} \end{cases}+ \mathrm{constant}$

The answer (Indefinite) [src]

                             /                 x            
                             |             1 - -            
                             |                 2            
                             |(2 - x - 2*y)           x     
                             |------------------  for - != 1
                             <          x             2     
  /                          |      1 - -                   
 |                           |          2                   
 |              -x           |                              
 |              ---          | log(2 - x - 2*y)   otherwise 
 |               2           \                              
 | (2 - x - 2*y)    dy = C - -------------------------------
 |                                          2               
/

$$\int \left(- 2 y + \left(2 - x\right)\right)^{\frac{\left(-1\right) x}{2}}\, dy = C - \frac{\begin{cases} \frac{\left(- 2 y + \left(2 - x\right)\right)^{1 - \frac{x}{2}}}{1 - \frac{x}{2}} & \text{for}\: \frac{x}{2} \neq 1 \\\log{\left(- 2 y + \left(2 - x\right) \right)} & \text{otherwise} \end{cases}}{2}$$

Simplify

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

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Identical expressions

Similar expressions

Integral of (2-x-2*y)^((-x)/2) dy

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3