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0-1-arccotgh-x-1-x-2-dx-by-M-A-




Question Number 163789 by amin96 last updated on 10/Jan/22
∫_0 ^1 ((arccotgh(x))/(1−x^2 ))dx=?  by M.A
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{{arccotgh}}\left(\boldsymbol{{x}}\right)}{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}}=? \\ $$$$\boldsymbol{{by}}\:\boldsymbol{{M}}.\boldsymbol{{A}} \\ $$
Commented by MJS_new last updated on 10/Jan/22
tanh x =((e^(2x) −1)/(e^(2x) +1))  coth x =((e^(2x) +1)/(e^(2x) −1))  arcoth x =(1/2)ln ((x+1)/(x−1))  (1/2)∫((ln ((x+1)/(x−1)))/(1−x^2 ))dx=(1/8)(ln ((x+1)/(x−1)))^2 +C  ⇒ ∫_0 ^1 ((arcoth x)/(1−x^2 ))dx doesn′t converge
$$\mathrm{tanh}\:{x}\:=\frac{\mathrm{e}^{\mathrm{2}{x}} −\mathrm{1}}{\mathrm{e}^{\mathrm{2}{x}} +\mathrm{1}} \\ $$$$\mathrm{coth}\:{x}\:=\frac{\mathrm{e}^{\mathrm{2}{x}} +\mathrm{1}}{\mathrm{e}^{\mathrm{2}{x}} −\mathrm{1}} \\ $$$$\mathrm{arcoth}\:{x}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{ln}\:\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{ln}\:\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} +{C} \\ $$$$\Rightarrow\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{arcoth}\:{x}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{converge} \\ $$

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