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Derivative of:
(x^2+x+2)/(x-1)
Identical expressions
(x^ two +x+ two)/(x- one)
(x squared plus x plus 2) divide by (x minus 1)
(x to the power of two plus x plus two) divide by (x minus one)
(x2+x+2)/(x-1)
x2+x+2/x-1
(x²+x+2)/(x-1)
(x to the power of 2+x+2)/(x-1)
x^2+x+2/x-1
(x^2+x+2) divide by (x-1)
(x^2+x+2)/(x-1)dx
Similar expressions
(x^2+x-2)/(x-1)
(x^2+x+2)/(x+1)
(x^2-x+2)/(x-1)

Integral of (x^2+x+2)/(x-1) dx

The solution

You have entered [src]

  0              
  /              
 |               
 |   2           
 |  x  + x + 2   
 |  ---------- dx
 |    x - 1      
 |               
/                
-1

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

Integral((x^2 + x + 2)/(x - 1), (x, -1, 0))

Detail solution

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

$\frac{x^{2}}{2} + 2 x + 4 \log{\left(x - 1 \right)}+ \mathrm{constant}$

The answer is:

$\frac{x^{2}}{2} + 2 x + 4 \log{\left(x - 1 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                            
 |                                             
 |  2                   2                      
 | x  + x + 2          x                       
 | ---------- dx = C + -- + 2*x + 4*log(-1 + x)
 |   x - 1             2                       
 |                                             
/

$$\int \frac{\left(x^{2} + x\right) + 2}{x - 1}\, dx = C + \frac{x^{2}}{2} + 2 x + 4 \log{\left(x - 1 \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

3/2 - 4*log(2)

$$\frac{3}{2} - 4 \log{\left(2 \right)}$$

3/2 - 4*log(2)

$$\frac{3}{2} - 4 \log{\left(2 \right)}$$

3/2 - 4*log(2)

Numerical answer [src]

-1.27258872223978

-1.27258872223978

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (x^2+x+2)/(x-1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2