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Integral of d{x}:
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Graphing y =:
(x²+1)/(x²-1)
Identical expressions
(x²+ one)/(x²- one)
(x² plus 1) divide by (x² minus 1)
(x² plus one) divide by (x² minus one)
x²+1/x²-1
(x²+1) divide by (x²-1)
(x²+1)/(x²-1)dx
Similar expressions
(x²+1)/(x²+1)
(x²-1)/(x²-1)

Integral of (x²+1)/(x²-1) dx

The solution

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

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

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

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$$-\log \left(x+1\right)+x+\log \left(x-1\right)$$

Simplify

The answer [src]

-oo - pi*I

$${\it \%a}$$

-oo - pi*I

$$-\infty - i \pi$$

Упростить

Numerical answer [src]

-43.7841039667738

-43.7841039667738

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

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Identical expressions

Similar expressions

Integral of (x²+1)/(x²-1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2