Интеграл 1/(tan(x)-x) (dx)

Вы ввели

  1              
  /              
 |               
 |      1        
 |  ---------- dx
 |  tan(x) - x   
 |               
/                
0                

$$\int_{0}^{1} \frac{1}{- x + \tan{\left (x \right )}}\, dx$$

Ответ

  1                    1              
  /                    /              
 |                    |               
 |      1             |      1        
 |  ---------- dx = - |  ---------- dx
 |  tan(x) - x        |  x - tan(x)   
 |                    |               
/                    /                
0                    0                

$$\int_{0}^{1}{{{1}\over{\tan x-x}}\;dx}$$

Ответ (Неопределённый)

  /                      /             
 |                      |              
 |     1                |     1        
 | ---------- dx = C -  | ---------- dx
 | tan(x) - x           | x - tan(x)   
 |                      |              
/                      /               

$$-{{4\,\int {{{\left(x^2-1\right)\,\sin \left(2\,x\right)+2\,x\, \cos \left(2\,x\right)}\over{\left(x^4+2\,x^2+1\right)\,\sin ^2 \left(2\,x\right)+\left(-4\,x^3-4\,x\right)\,\sin \left(2\,x\right)+ \left(x^4+2\,x^2+1\right)\,\cos ^2\left(2\,x\right)+\left(2\,x^4-2 \right)\,\cos \left(2\,x\right)+x^4+2\,x^2+1}}}{\;dx}+\log \left(x^2 +1\right)}\over{2}}$$