Integral of (x+3)/(x-1) dx

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  1         
  /         
 |          
 |  x + 3   
 |  ----- dx
 |  x - 1   
 |          
/           
0

$$\int\limits_{0}^{1} \frac{x + 3}{x - 1}\, dx$$

Integral((x + 3)/(x - 1), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{x + 3}{x - 1} = 1 + \frac{4}{x - 1}$
2. Integrate term-by-term:
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 1\, dx = x$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{4}{x - 1}\, dx = 4 \int \frac{1}{x - 1}\, dx$
    1. Let $u = x - 1$ .
      
      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:
      
      $\log{\left(x - 1 \right)}$
    So, the result is: $4 \log{\left(x - 1 \right)}$
  The result is: $x + 4 \log{\left(x - 1 \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x + 3}{x - 1} = \frac{x}{x - 1} + \frac{3}{x - 1}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{x}{x - 1} = 1 + \frac{1}{x - 1}$
  2. Integrate term-by-term:
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
    1. Let $u = x - 1$ .
      
      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:
      
      $\log{\left(x - 1 \right)}$
    The result is: $x + \log{\left(x - 1 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{3}{x - 1}\, dx = 3 \int \frac{1}{x - 1}\, dx$
    1. Let $u = x - 1$ .
      
      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:
      
      $\log{\left(x - 1 \right)}$
    So, the result is: $3 \log{\left(x - 1 \right)}$
  The result is: $x + \log{\left(x - 1 \right)} + 3 \log{\left(x - 1 \right)}$
Add the constant of integration:

$x + 4 \log{\left(x - 1 \right)}+ \mathrm{constant}$

The answer is:

$x + 4 \log{\left(x - 1 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                
 |                                 
 | x + 3                           
 | ----- dx = C + x + 4*log(-1 + x)
 | x - 1                           
 |                                 
/

$$\int \frac{x + 3}{x - 1}\, dx = C + x + 4 \log{\left(x - 1 \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-oo - 4*pi*I

$$-\infty - 4 i \pi$$

=

-oo - 4*pi*I

$$-\infty - 4 i \pi$$

-oo - 4*pi*i

Numerical answer [src]

-175.363827144878

-175.363827144878

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Integral of (x+3)/(x-1) dx

Limits of integration:

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Piecewise:

The solution

Method #1

Method #2