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Identical expressions
arctgx/x^ two + one
arctgx divide by x squared plus 1
arctgx divide by x to the power of two plus one
arctgx/x2+1
arctgx/x²+1
arctgx/x to the power of 2+1
arctgx divide by x^2+1
arctgx/x^2+1dx
Similar expressions
arctgx/x^2-1
What you mean?
acot(x)/(x^2 + 1)
acot(x)/(x^2) + 1
acot(x)*2/x + 1
acot(x)/(x^2) + 1
acot(x)/(x^2 + 1)
acot(x)/(x^2) + 1
acot(x)/(x^2) + 1

You entered:

arctgx/x^2+1

What you mean?

acot(x)/(x^2 + 1)

acot(x)/(x^2) + 1

acot(x)*2/x + 1

acot(x)/(x^2) + 1

acot(x)/(x^2 + 1)

acot(x)/(x^2) + 1

acot(x)/(x^2) + 1

Integral of arctgx/x^2+1 dx

The solution

You have entered [src]

 oo                 
  /                 
 |                  
 |  /acot(x)    \   
 |  |------- + 1| dx
 |  |    2      |   
 |  \   x       /   
 |                  
/                   
0

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

Integral(acot(x)/(x^2) + 1, (x, 0, oo))

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The answer [src]

oo

$$\infty$$

oo

$$\infty$$

Упростить

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

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What you mean?

You entered:

What you mean?

Integral of arctgx/x^2+1 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4