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Integral of zdz/(z+2)x(z-i)2 dz

The solution

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 -3                     
  /                     
 |                      
 |    z                 
 |  -----*x*(z - I)*2 dz
 |  z + 2               
 |                      
/                       
0

$$\int\limits_{0}^{-3} 2 x \frac{z}{z + 2} \left(z - i\right)\, dz$$

Integral((((z/(z + 2))*x)*(z - i))*2, (z, 0, -3))

Detail solution

The integral of a constant times a function is the constant times the integral of the function:

$\int 2 x \frac{z}{z + 2} \left(z - i\right)\, dz = 2 \int \frac{x z \left(z - i\right)}{z + 2}\, dz$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{x z \left(z - i\right)}{z + 2}\, dz = x \int \frac{z \left(z - i\right)}{z + 2}\, dz$
  1. There are multiple ways to do this integral.
    Method #1
    1. Rewrite the integrand:
      
      $\frac{z \left(z - i\right)}{z + 2} = z - 2 - i + \frac{2 \left(2 + i\right)}{z + 2}$
    2. Integrate term-by-term:
      
      The integral of $z^{n}$ is $\frac{z^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int z\, dz = \frac{z^{2}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-2\right)\, dz = - 2 z$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(- i\right)\, dz = - i z$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{2 \left(2 + i\right)}{z + 2}\, dz = \left(4 + 2 i\right) \int \frac{1}{z + 2}\, dz$
      
      Let $u = z + 2$ .
      
      Then let $du = dz$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(z + 2 \right)}$
      
      So, the result is: $\left(4 + 2 i\right) \log{\left(z + 2 \right)}$
      
      The result is: $\frac{z^{2}}{2} - 2 z - i z + \left(4 + 2 i\right) \log{\left(z + 2 \right)}$
    Method #2
    1. Rewrite the integrand:
      
      $\frac{z \left(z - i\right)}{z + 2} = \frac{z^{2} - i z}{z + 2}$
    2. Rewrite the integrand:
      
      $\frac{z^{2} - i z}{z + 2} = z - 2 - i + \frac{2 \left(2 + i\right)}{z + 2}$
    3. Integrate term-by-term:
      
      The integral of $z^{n}$ is $\frac{z^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int z\, dz = \frac{z^{2}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-2\right)\, dz = - 2 z$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(- i\right)\, dz = - i z$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{2 \left(2 + i\right)}{z + 2}\, dz = \left(4 + 2 i\right) \int \frac{1}{z + 2}\, dz$
      
      Let $u = z + 2$ .
      
      Then let $du = dz$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(z + 2 \right)}$
      
      So, the result is: $\left(4 + 2 i\right) \log{\left(z + 2 \right)}$
      
      The result is: $\frac{z^{2}}{2} - 2 z - i z + \left(4 + 2 i\right) \log{\left(z + 2 \right)}$
    Method #3
    1. Rewrite the integrand:
      
      $\frac{z \left(z - i\right)}{z + 2} = \frac{z^{2}}{z + 2} - \frac{i z}{z + 2}$
    2. Integrate term-by-term:
      
      Rewrite the integrand:
      
      $\frac{z^{2}}{z + 2} = z - 2 + \frac{4}{z + 2}$
      
      Integrate term-by-term:
      
      The integral of $z^{n}$ is $\frac{z^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int z\, dz = \frac{z^{2}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-2\right)\, dz = - 2 z$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{4}{z + 2}\, dz = 4 \int \frac{1}{z + 2}\, dz$
      
      Let $u = z + 2$ .
      
      Then let $du = dz$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(z + 2 \right)}$
      
      So, the result is: $4 \log{\left(z + 2 \right)}$
      
      The result is: $\frac{z^{2}}{2} - 2 z + 4 \log{\left(z + 2 \right)}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{i z}{z + 2}\right)\, dz = - i \int \frac{z}{z + 2}\, dz$
      
      Rewrite the integrand:
      
      $\frac{z}{z + 2} = 1 - \frac{2}{z + 2}$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dz = z$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{2}{z + 2}\right)\, dz = - 2 \int \frac{1}{z + 2}\, dz$
      
      Let $u = z + 2$ .
      
      Then let $du = dz$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(z + 2 \right)}$
      
      So, the result is: $- 2 \log{\left(z + 2 \right)}$
      
      The result is: $z - 2 \log{\left(z + 2 \right)}$
      
      So, the result is: $- i \left(z - 2 \log{\left(z + 2 \right)}\right)$
      
      The result is: $\frac{z^{2}}{2} - 2 z - i \left(z - 2 \log{\left(z + 2 \right)}\right) + 4 \log{\left(z + 2 \right)}$
  So, the result is: $x \left(\frac{z^{2}}{2} - 2 z - i z + \left(4 + 2 i\right) \log{\left(z + 2 \right)}\right)$
So, the result is: $2 x \left(\frac{z^{2}}{2} - 2 z - i z + \left(4 + 2 i\right) \log{\left(z + 2 \right)}\right)$
Now simplify:

$x \left(z^{2} - 4 z - 2 i z + 4 \left(2 + i\right) \log{\left(z + 2 \right)}\right)$
Add the constant of integration:

$x \left(z^{2} - 4 z - 2 i z + 4 \left(2 + i\right) \log{\left(z + 2 \right)}\right)+ \mathrm{constant}$

The answer is:

$x \left(z^{2} - 4 z - 2 i z + 4 \left(2 + i\right) \log{\left(z + 2 \right)}\right)+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                      
 |                                / 2                                   \
 |   z                            |z                                    |
 | -----*x*(z - I)*2 dz = C + 2*x*|-- - 2*z + (4 + 2*I)*log(2 + z) - I*z|
 | z + 2                          \2                                    /
 |                                                                       
/

$$\int 2 x \frac{z}{z + 2} \left(z - i\right)\, dz = C + 2 x \left(\frac{z^{2}}{2} - 2 z - i z + \left(4 + 2 i\right) \log{\left(z + 2 \right)}\right)$$

Simplify

The answer [src]

nan

$$\text{NaN}$$

nan

$$\text{NaN}$$

nan

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

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Identical expressions

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Integral of zdz/(z+2)x(z-i)2 dz

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3