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Integral of 1/(x(1+x^2))
Identical expressions
one /(x(one +x^ two))
1 divide by (x(1 plus x squared ))
one divide by (x(one plus x to the power of two))
1/(x(1+x2))
1/x1+x2
1/(x(1+x²))
1/(x(1+x to the power of 2))
1/x1+x^2
1 divide by (x(1+x^2))
1/(x(1+x^2))dx
Similar expressions
1/(x(1-x^2))

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

The solution

You have entered [src]

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

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

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

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$$\log x-{{\log \left(x^2+1\right)}\over{2}}$$

Simplify

The answer [src]

oo

$${\it \%a}$$

oo

$$\infty$$

Упростить

Numerical answer [src]

43.7438725437129

43.7438725437129

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3