Question Number 221587 by Tawa11 last updated on 08/Jun/25
$$\int_{\:\mathrm{2}} ^{\:\mathrm{3}} \:\frac{\mathrm{tan}^{−\:\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{1}\:\:−\:\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$
Answered by maths2 last updated on 09/Jun/25
![=∫_2 ^3 (1/2)[(1/(1+x))+(1/(1−x))]tan^(−1) (x)dx=Z =(1/2)ln(∣((1+x)/(1−x))∣]_2 ^3 tan^(−1) (x)−(1/2)∫_2 ^3 ((ln(((1+x)/(1−x))))/(1+x^2 ))dx .((ln(2)tan^(−1) (3)−ln(3)tan^(−1) (2))/2)−(1/2)A A;((1+x)/(1−x))=u⇒x=((u−1)/(1+u))⇒dx=(2/((1+u)^2 ))du A=∫_3 ^2 ((ln(u))/(u^2 +1))du=^(u→(1/u)) ∫_(1/3) ^(1/2) ((ln(u))/(1+u^2 ))du=∫_(1/3) ^(1/2) (1/2)((1/(1+iu))+(1/(1−iu)))ln(u)du =Re∫_(1/3) ^(1/2) ((ln(u))/(1+iu))du=Re[−iln(1+iu)ln(u)]+Re{∫_(1/3) ^(1/2) i((ln(1+iu))/u)du} =−Re(iln(1+(i/2))ln((1/2))−iln(1+(i/3))ln((1/3))) −Re{−∫_(1/3) ^(1/2) i((ln(1−(−iu)))/(−iu))d(−iu)}_(=B) B=−Re{Li_2 (−iu)]_(1/3) ^(1/2) A=−Re(iln(1+(i/2))ln((1/2))−iln(1+(i/3))ln((1/3))] −Re(Li_2 (−(i/2))−Li_2 (−(i/3))) Z=((ln(2)tan^(−1) (3)−ln(3)tan^(−1) (2))/2)+(1/2)Re(iln(1+(i/2))ln((1/2))−iln(1+(i/3))ln((1/3))+Li_2 (−(i/2))−Li_2 (−(i/3))) Re reel part Li_2 (z)=∫_0 ^z −((ln(1−g))/g)dg](https://www.tinkutara.com/question/Q221667.png)
$$=\int_{\mathrm{2}} ^{\mathrm{3}} \frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{1}−{x}}\right]\mathrm{tan}^{−\mathrm{1}} \left({x}\right){dx}={Z} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mid\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\mid\right]_{\mathrm{2}} ^{\mathrm{3}} \mathrm{tan}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{2}} ^{\mathrm{3}} \frac{{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$.\frac{{ln}\left(\mathrm{2}\right)\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}\right)−{ln}\left(\mathrm{3}\right)\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{A} \\ $$$${A};\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}={u}\Rightarrow{x}=\frac{{u}−\mathrm{1}}{\mathrm{1}+{u}}\Rightarrow{dx}=\frac{\mathrm{2}}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }{du} \\ $$$${A}=\int_{\mathrm{3}} ^{\mathrm{2}} \frac{{ln}\left({u}\right)}{{u}^{\mathrm{2}} +\mathrm{1}}{du}\overset{{u}\rightarrow\frac{\mathrm{1}}{{u}}} {=}\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{ln}\left({u}\right)}{\mathrm{1}+{u}^{\mathrm{2}} }{du}=\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}+{iu}}+\frac{\mathrm{1}}{\mathrm{1}−{iu}}\right){ln}\left({u}\right){du} \\ $$$$={Re}\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{ln}\left({u}\right)}{\mathrm{1}+{iu}}{du}={Re}\left[−{iln}\left(\mathrm{1}+{iu}\right){ln}\left({u}\right)\right]+{Re}\left\{\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {i}\frac{{ln}\left(\mathrm{1}+{iu}\right)}{{u}}{du}\right\} \\ $$$$=−{Re}\left({iln}\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right){ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−{iln}\left(\mathrm{1}+\frac{{i}}{\mathrm{3}}\right){ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right) \\ $$$$−{Re}\left\{−\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {i}\frac{{ln}\left(\mathrm{1}−\left(−{iu}\right)\right)}{−{iu}}{d}\left(−{iu}\right)\right\}_{={B}} \\ $$$${B}=−{Re}\left\{{Li}_{\mathrm{2}} \left(−{iu}\right)\right]_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${A}=−{Re}\left({iln}\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right){ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−{iln}\left(\mathrm{1}+\frac{{i}}{\mathrm{3}}\right){ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right] \\ $$$$−{Re}\left({Li}_{\mathrm{2}} \left(−\frac{{i}}{\mathrm{2}}\right)−{Li}_{\mathrm{2}} \left(−\frac{{i}}{\mathrm{3}}\right)\right) \\ $$$${Z}=\frac{{ln}\left(\mathrm{2}\right)\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}\right)−{ln}\left(\mathrm{3}\right)\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{Re}\left({iln}\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right){ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−{iln}\left(\mathrm{1}+\frac{{i}}{\mathrm{3}}\right){ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+{Li}_{\mathrm{2}} \left(−\frac{{i}}{\mathrm{2}}\right)−{Li}_{\mathrm{2}} \left(−\frac{{i}}{\mathrm{3}}\right)\right) \\ $$$${Re}\:{reel}\:{part} \\ $$$${Li}_{\mathrm{2}} \left({z}\right)=\int_{\mathrm{0}} ^{{z}} −\frac{{ln}\left(\mathrm{1}−{g}\right)}{{g}}{dg} \\ $$$$ \\ $$
Commented by Tawa11 last updated on 09/Jun/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{computation}\:\mathrm{for}\:\mathrm{the}\:\mathrm{final}\:\mathrm{answer}. \\ $$