Integral of (x+b)÷(x+c)dx dx

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  1         
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 |  x + b   
 |  ----- dx
 |  x + c   
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0

$$\int\limits_{0}^{1} \frac{b + x}{c + x}\, dx$$

Integral((x + b)/(x + c), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{b + x}{c + x} = \frac{b - c}{c + x} + 1$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{b - c}{c + x}\, dx = \left(b - c\right) \int \frac{1}{c + x}\, dx$
    1. Let $u = c + x$ .
      
      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:
      
      $\log{\left(c + x \right)}$
    So, the result is: $\left(b - c\right) \log{\left(c + x \right)}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 1\, dx = x$
  The result is: $x + \left(b - c\right) \log{\left(c + x \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{b + x}{c + x} = \frac{b}{c + x} + \frac{x}{c + x}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{b}{c + x}\, dx = b \int \frac{1}{c + x}\, dx$
    1. Let $u = c + x$ .
      
      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:
      
      $\log{\left(c + x \right)}$
    So, the result is: $b \log{\left(c + x \right)}$
  1. Rewrite the integrand:
    
    $\frac{x}{c + x} = - \frac{c}{c + x} + 1$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{c}{c + x}\right)\, dx = - c \int \frac{1}{c + x}\, dx$
      
      Let $u = c + x$ .
      
      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:
      
      $\log{\left(c + x \right)}$
      
      So, the result is: $- c \log{\left(c + x \right)}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
    The result is: $- c \log{\left(c + x \right)} + x$
  The result is: $b \log{\left(c + x \right)} - c \log{\left(c + x \right)} + x$
Add the constant of integration:

$x + \left(b - c\right) \log{\left(c + x \right)}+ \mathrm{constant}$

The answer is:

$x + \left(b - c\right) \log{\left(c + x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                     
 |                                      
 | x + b                                
 | ----- dx = C + x + (b - c)*log(c + x)
 | x + c                                
 |                                      
/

$$\int \frac{b + x}{c + x}\, dx = C + x + \left(b - c\right) \log{\left(c + x \right)}$$

Simplify

The answer [src]

1 + (b - c)*log(1 + c) - (b - c)*log(c)

$$- \left(b - c\right) \log{\left(c \right)} + \left(b - c\right) \log{\left(c + 1 \right)} + 1$$

=

1 + (b - c)*log(1 + c) - (b - c)*log(c)

$$- \left(b - c\right) \log{\left(c \right)} + \left(b - c\right) \log{\left(c + 1 \right)} + 1$$

1 + (b - c)*log(1 + c) - (b - c)*log(c)

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Integral of (x+b)÷(x+c)dx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2