Integral of dx/(x+a)(x+b) dx

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

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

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

Detail solution

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

$x - \left(a - b\right) \log{\left(a + x \right)}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The answer [src]

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

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

=

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

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

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

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Integral of dx/(x+a)(x+b) dx

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The graph:

Piecewise:

The solution

Method #1

Method #2