Интеграл acot(x)^2 (dx)

Вы ввели

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 |      2      
 |  acot (x) dx
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0              

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

Ответ

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 |  acot (x) dx =  |  acot (x) dx
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0                 0              

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

Численный ответ

1.33407424861857

Ответ (Неопределённый)

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