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Solving integrals/ arcctg(x)^9

Integral of arcctg(x)^9 dx

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∫01acot⁡9(x) dx\int\limits_{0}^{1} \operatorname{acot}^{9}{\left(x \right)}\, dx 
Integral(acot(x)^9, (x, 0, 1))
The answer (Indefinite) [src]
−131072 (9 ∫arctan⁡(1x) x2 (log⁡(x2+1))8131072 x2+131072  dx−9 ∫x (log⁡(x2+1))8131072 x2+131072  dx+9 ∫arctan⁡(1x) (log⁡(x2+1))8131072 x2+131072  dx+144 ∫arctan⁡(1x) x2 (log⁡(x2+1))7131072 x2+131072  dx−336 ∫arctan⁡3(1x) x2 (log⁡(x2+1))6131072 x2+131072  dx+1008 ∫arctan⁡2(1x) x (log⁡(x2+1))6131072 x2+131072  dx−336 ∫arctan⁡3(1x) (log⁡(x2+1))6131072 x2+131072  dx−4032 ∫arctan⁡3(1x) x2 (log⁡(x2+1))5131072 x2+131072  dx+2016 ∫arctan⁡5(1x) x2 (log⁡(x2+1))4131072 x2+131072  dx−10080 ∫arctan⁡4(1x) x (log⁡(x2+1))4131072 x2+131072  dx+2016 ∫arctan⁡5(1x) (log⁡(x2+1))4131072 x2+131072  dx+16128 ∫arctan⁡5(1x) x2 (log⁡(x2+1))3131072 x2+131072  dx−2304 ∫arctan⁡7(1x) x2 (log⁡(x2+1))2131072 x2+131072  dx+16128 ∫arctan⁡6(1x) x (log⁡(x2+1))2131072 x2+131072  dx−2304 ∫arctan⁡7(1x) (log⁡(x2+1))2131072 x2+131072  dx−9216 ∫arctan⁡7(1x) x2 log⁡(x2+1)131072 x2+131072  dx−130816 ∫arctan⁡9(1x) x2131072 x2+131072  dx−2304 ∫arctan⁡8(1x) x131072 x2+131072  dx+130816 (9 (4 (7 (3 (2 (3 (−arctan⁡10x360−arctan⁡(1x) arctan⁡9x36)7−3 arctan⁡2(1x) arctan⁡8x56)3−2 arctan⁡3(1x) arctan⁡7x21−arctan⁡4(1x) arctan⁡6x6)2−3 arctan⁡5(1x) arctan⁡5x10)3−7 arctan⁡6(1x) arctan⁡4x12)−4 arctan⁡7(1x) arctan⁡3x3)131072−9 arctan⁡8(1x) arctan⁡2x262144)−511 arctan⁡9(1x) arctan⁡x512)−9 atan2(1,x) x (log⁡(x2+1))8+336 atan2(1,x)3 x (log⁡(x2+1))6−2016 atan2(1,x)5 x (log⁡(x2+1))4+2304 atan2(1,x)7 x (log⁡(x2+1))2−256 atan2(1,x)9 x131072-{{131072\,\left(9\,\int {{{\arctan \left({{1}\over{x}}\right)\,x^2 \,\left(\log \left(x^2+1\right)\right)^8}\over{131072\,x^2+131072}} }{\;dx}-9\,\int {{{x\,\left(\log \left(x^2+1\right)\right)^8}\over{ 131072\,x^2+131072}}}{\;dx}+9\,\int {{{\arctan \left({{1}\over{x}} \right)\,\left(\log \left(x^2+1\right)\right)^8}\over{131072\,x^2+ 131072}}}{\;dx}+144\,\int {{{\arctan \left({{1}\over{x}}\right)\,x^2 \,\left(\log \left(x^2+1\right)\right)^7}\over{131072\,x^2+131072}} }{\;dx}-336\,\int {{{\arctan ^3\left({{1}\over{x}}\right)\,x^2\, \left(\log \left(x^2+1\right)\right)^6}\over{131072\,x^2+131072}} }{\;dx}+1008\,\int {{{\arctan ^2\left({{1}\over{x}}\right)\,x\, \left(\log \left(x^2+1\right)\right)^6}\over{131072\,x^2+131072}} }{\;dx}-336\,\int {{{\arctan ^3\left({{1}\over{x}}\right)\,\left( \log \left(x^2+1\right)\right)^6}\over{131072\,x^2+131072}}}{\;dx}- 4032\,\int {{{\arctan ^3\left({{1}\over{x}}\right)\,x^2\,\left(\log \left(x^2+1\right)\right)^5}\over{131072\,x^2+131072}}}{\;dx}+2016\, \int {{{\arctan ^5\left({{1}\over{x}}\right)\,x^2\,\left(\log \left( x^2+1\right)\right)^4}\over{131072\,x^2+131072}}}{\;dx}-10080\, \int {{{\arctan ^4\left({{1}\over{x}}\right)\,x\,\left(\log \left(x^ 2+1\right)\right)^4}\over{131072\,x^2+131072}}}{\;dx}+2016\,\int {{{ \arctan ^5\left({{1}\over{x}}\right)\,\left(\log \left(x^2+1\right) \right)^4}\over{131072\,x^2+131072}}}{\;dx}+16128\,\int {{{\arctan ^ 5\left({{1}\over{x}}\right)\,x^2\,\left(\log \left(x^2+1\right) \right)^3}\over{131072\,x^2+131072}}}{\;dx}-2304\,\int {{{\arctan ^7 \left({{1}\over{x}}\right)\,x^2\,\left(\log \left(x^2+1\right) \right)^2}\over{131072\,x^2+131072}}}{\;dx}+16128\,\int {{{\arctan ^ 6\left({{1}\over{x}}\right)\,x\,\left(\log \left(x^2+1\right)\right) ^2}\over{131072\,x^2+131072}}}{\;dx}-2304\,\int {{{\arctan ^7\left( {{1}\over{x}}\right)\,\left(\log \left(x^2+1\right)\right)^2}\over{ 131072\,x^2+131072}}}{\;dx}-9216\,\int {{{\arctan ^7\left({{1}\over{ x}}\right)\,x^2\,\log \left(x^2+1\right)}\over{131072\,x^2+131072}} }{\;dx}-130816\,\int {{{\arctan ^9\left({{1}\over{x}}\right)\,x^2 }\over{131072\,x^2+131072}}}{\;dx}-2304\,\int {{{\arctan ^8\left({{1 }\over{x}}\right)\,x}\over{131072\,x^2+131072}}}{\;dx}+130816\, \left({{9\,\left(4\,\left({{7\,\left({{3\,\left({{2\,\left({{3\, \left(-{{\arctan ^{10}x}\over{360}}-{{\arctan \left({{1}\over{x}} \right)\,\arctan ^9x}\over{36}}\right)}\over{7}}-{{3\,\arctan ^2 \left({{1}\over{x}}\right)\,\arctan ^8x}\over{56}}\right)}\over{3}}- {{2\,\arctan ^3\left({{1}\over{x}}\right)\,\arctan ^7x}\over{21}}-{{ \arctan ^4\left({{1}\over{x}}\right)\,\arctan ^6x}\over{6}}\right) }\over{2}}-{{3\,\arctan ^5\left({{1}\over{x}}\right)\,\arctan ^5x }\over{10}}\right)}\over{3}}-{{7\,\arctan ^6\left({{1}\over{x}} \right)\,\arctan ^4x}\over{12}}\right)-{{4\,\arctan ^7\left({{1 }\over{x}}\right)\,\arctan ^3x}\over{3}}\right)}\over{131072}}-{{9\, \arctan ^8\left({{1}\over{x}}\right)\,\arctan ^2x}\over{262144}} \right)-{{511\,\arctan ^9\left({{1}\over{x}}\right)\,\arctan x }\over{512}}\right)-9\,{\rm atan2}\left(1 , x\right)\,x\,\left(\log \left(x^2+1\right)\right)^8+336\,{\rm atan2}\left(1 , x\right)^3\,x \,\left(\log \left(x^2+1\right)\right)^6-2016\,{\rm atan2}\left(1 , x\right)^5\,x\,\left(\log \left(x^2+1\right)\right)^4+2304\, {\rm atan2}\left(1 , x\right)^7\,x\,\left(\log \left(x^2+1\right) \right)^2-256\,{\rm atan2}\left(1 , x\right)^9\,x}\over{131072}}
Simplify
The answer [src] 
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∫01(arccot  x)9  dx\int_{0}^{1}{\left({\rm arccot}\; x\right)^9\;dx} 
=
= 
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∫01acot⁡9(x) dx\int\limits_{0}^{1} \operatorname{acot}^{9}{\left(x \right)}\, dx
Упростить
Numerical answer [src] 
9.51158422982041
9.51158422982041

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

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