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Identical expressions
arcctg^ two x/ one +x^2
arcctg squared x divide by 1 plus x squared
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arcctg2x/1+x2
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arcctg^2x divide by 1+x^2
arcctg^2x/1+x^2dx
Similar expressions
arcctg^2x/1-x^2

Integral of arcctg^2x/1+x^2 dx

The solution

You have entered [src]

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

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

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

Detail solution

Integrate term-by-term:
1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
  
  $\int x^{2}\, dx = \frac{x^{3}}{3}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{\operatorname{acot}^{2}{\left(x \right)}}{1}\, dx = \int \operatorname{acot}^{2}{\left(x \right)}\, dx$
  1. Don't know the steps in finding this integral.
    
    But the integral is
    
    $\int \operatorname{acot}^{2}{\left(x \right)}\, dx$
  So, the result is: $\int \operatorname{acot}^{2}{\left(x \right)}\, dx$
The result is: $\frac{x^{3}}{3} + \int \operatorname{acot}^{2}{\left(x \right)}\, dx$
Add the constant of integration:

$\frac{x^{3}}{3} + \int \operatorname{acot}^{2}{\left(x \right)}\, dx+ \mathrm{constant}$

The answer is:

$\frac{x^{3}}{3} + \int \operatorname{acot}^{2}{\left(x \right)}\, dx+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                            
 |                                 /           
 | /    2        \           3    |            
 | |acot (x)    2|          x     |     2      
 | |-------- + x | dx = C + -- +  | acot (x) dx
 | \   1         /          3     |            
 |                               /             
/

$${{16\,\left(\int {{{x^2\,\left(\log \left(x^2+1\right)\right)^2 }\over{16\,x^2+16}}}{\;dx}+\int {{{\left(\log \left(x^2+1\right) \right)^2}\over{16\,x^2+16}}}{\;dx}+4\,\int {{{x^2\,\log \left(x^2+1 \right)}\over{16\,x^2+16}}}{\;dx}+12\,\int {{{\arctan ^2\left({{1 }\over{x}}\right)\,x^2}\over{16\,x^2+16}}}{\;dx}+8\,\int {{{\arctan \left({{1}\over{x}}\right)\,x}\over{16\,x^2+16}}}{\;dx}-12\,\left(- {{\arctan ^3x}\over{48}}-{{\arctan \left({{1}\over{x}}\right)\, \arctan ^2x}\over{16}}\right)+{{3\,\arctan ^2\left({{1}\over{x}} \right)\,\arctan x}\over{4}}\right)-x\,\left(\log \left(x^2+1\right) \right)^2+4\,{\rm atan2}\left(1 , x\right)^2\,x}\over{16}}+{{x^3 }\over{3}}$$

The answer [src]

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

$$\int_{0}^{1}{\left({\rm arccot}\; x\right)^2+x^2\;dx}$$

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

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

Simplify

Numerical answer [src]

1.6674075819519

1.6674075819519

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

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Integral of arcctg^2x/1+x^2 dx

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The graph:

Piecewise:

The solution