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Integral of d{x}:
Integral of 2/x^4
Integral of e^(3*x)/3
Integral of log(x+1)/(x^2+1)
Integral of 1/(x^3-1)^2
Graphing y =:
(x+1)/(x^2-4)
Identical expressions
(x+ one)/(x^ two - four)
(x plus 1) divide by (x squared minus 4)
(x plus one) divide by (x to the power of two minus four)
(x+1)/(x2-4)
x+1/x2-4
(x+1)/(x²-4)
(x+1)/(x to the power of 2-4)
x+1/x^2-4
(x+1) divide by (x^2-4)
(x+1)/(x^2-4)dx
Similar expressions
(x+1)/(x^2+4)
(x-1)/(x^2-4)

Solving integrals/ (x+1)/(x^2-4)

Integral of (x+1)/(x^2-4) dx

The solution

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  1          
  /          
 |           
 |  x + 1    
 |  ------ dx
 |   2       
 |  x  - 4   
 |           
/            
0

$$\int\limits_{0}^{1} \frac{x + 1}{x^{2} - 4}\, dx$$

Integral((x + 1)/(x^2 - 1*4), (x, 0, 1))

Detail solution

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

$\frac{3 \log{\left(x - 2 \right)}}{4} + \frac{\log{\left(x + 2 \right)}}{4}+ \mathrm{constant}$

The answer is:

$\frac{3 \log{\left(x - 2 \right)}}{4} + \frac{\log{\left(x + 2 \right)}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                          
 |                                           
 | x + 1           log(2 + x)   3*log(-2 + x)
 | ------ dx = C + ---------- + -------------
 |  2                  4              4      
 | x  - 4                                    
 |                                           
/

$${{\log \left(x+2\right)}\over{4}}+{{3\,\log \left(x-2\right)}\over{ 4}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

          log(3)
-log(2) + ------
            4

$${{\log 3}\over{4}}-\log 2$$

          log(3)
-log(2) + ------
            4

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

Упростить

Numerical answer [src]

-0.418494108392918

-0.418494108392918

The graph

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

Other calculators

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Identical expressions

Similar expressions

Integral of (x+1)/(x^2-4) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2