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Question Number 72665 by TawaTawa last updated on 31/Oct/19
∫_( 2) ^( 3) ((tan^(−1) (x))/(1 − x^2 )) dx
$$\int_{\:\mathrm{2}} ^{\:\mathrm{3}} \:\:\frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{1}\:−\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$
Commented by mind is power last updated on 31/Oct/19
non close value we have to use Li_2 (x) function
$$\mathrm{non}\:\mathrm{close}\:\mathrm{value}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\:\mathrm{use}\:\mathrm{Li}_{\mathrm{2}} \left(\mathrm{x}\right)\:\mathrm{function}\: \\ $$$$ \\ $$
Commented by TawaTawa last updated on 31/Oct/19
Help sir
$$\mathrm{Help}\:\mathrm{sir} \\ $$
Commented by mathmax by abdo last updated on 31/Oct/19
let I =∫_2 ^3 ((arctan(x))/(1−x^2 ))dx =_(x=2+t) ∫_0 ^1 ((arctan(2+t))/(1−(2+t)^2 ))dt =−∫_0 ^1 ((arctan(2+t))/((t+2)^2 −1))dt =−∫_0 ^1 ((arctan(t+2))/((t+1)(t+3)))dt =−(1/2)∫_0 ^1 arctan(t+2)((1/(t+1))−(1/(t+3)))dt =(1/2)∫_0 ^1 ((arctan(t+2))/(t+3))dt−(1/2)∫_0 ^1 ((arctan(t+2))/(t+1))dt =(1/2)H−(1/2)K H=∫_0 ^1 ((arctan(t+2))/(t+3))dt and K=∫_0 ^1 ((arctan(t+2))/(t+1))dt let f(x)=∫_0 ^1 ((arctan(t+x))/(t+3))dt with x≥0 ⇒ f^′ (x)=∫_0 ^1 (1/((t+3)(1+(t+x)^2 )))dt =∫_0 ^1 (dt/((t+3)(t^2 +2xt +x^2 +1))) let decompose F(t)=(1/((t+3)(t^2 +2xt +x^2 +1))) F(t)=(a/(t+3)) +((bt +c)/(t^2 +2xt +x^2 +1)) a =(1/(10−6x +x^2 )) lim_(t→+∞) tF(t)=0=a+b ⇒b=((−1)/(x^2 −6x +10)) F(0)=(1/(3(x^2 +1))) =(a/3) +(c/(x^2 +1)) ⇒(1/3) =((x^2 +1)/3)a +c ⇒ 1=(x^2 +1)a +3c ⇒3c=1−(x^2 +1)×(1/(x^2 −6x +10)) ⇒c =(1/3)(1−((x^2 +1)/(x^2 −6x +10))) ∫_0 ^1 F(t)dt =a ∫_0 ^1 (dt/(t+3)) +(b/2) ∫_0 ^1 ((2t +2x−2x )/(t^2 +2xt +x^2 +1))dt =a[ln(t+3)]_0 ^1 +(b/2)ln(t^2 +2xt +x^2 +1)]_0 ^x +c ∫_0 ^1 (dt/(t^2 +2xt +x^2 +1)) ∫_0 ^1 (dt/(t^2 +2xt +x^2 +1)) =∫_0 ^1 (dt/((t+x)^2 +1)) =_(t+x=u) ∫_x ^(x+1) (du/(1+u^2 )) =arctan(x+1)−arctanx ⇒ f^′ (x)=a{ln((4/3))}+(b/2)(ln(4x^2 +1)−ln(x^2 +1)+c(arctan(x+1)−arctanx) ⇒f(x)=aln((4/3))x +(b/2)∫ln(((4x^2 +1)/(x^2 +1)))dx +c∫ (arctan(x+1)−arctan(x))dx +c ....be continued....
$${let}\:{I}\:=\int_{\mathrm{2}} ^{\mathrm{3}} \:\frac{{arctan}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:=_{{x}=\mathrm{2}+{t}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left(\mathrm{2}+{t}\right)}{\mathrm{1}−\left(\mathrm{2}+{t}\right)^{\mathrm{2}} }{dt} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{arctan}\left(\mathrm{2}+{t}\right)}{\left({t}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}}{dt}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({t}+\mathrm{2}\right)}{\left({t}+\mathrm{1}\right)\left({t}+\mathrm{3}\right)}{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:{arctan}\left({t}+\mathrm{2}\right)\left(\frac{\mathrm{1}}{{t}+\mathrm{1}}−\frac{\mathrm{1}}{{t}+\mathrm{3}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({t}+\mathrm{2}\right)}{{t}+\mathrm{3}}{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({t}+\mathrm{2}\right)}{{t}+\mathrm{1}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{H}−\frac{\mathrm{1}}{\mathrm{2}}{K}\:\: \\ $$$${H}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{arctan}\left({t}+\mathrm{2}\right)}{{t}+\mathrm{3}}{dt}\:{and}\:{K}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{arctan}\left({t}+\mathrm{2}\right)}{{t}+\mathrm{1}}{dt} \\ $$$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{arctan}\left({t}+{x}\right)}{{t}+\mathrm{3}}{dt}\:\:{with}\:{x}\geqslant\mathrm{0}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{\left({t}+\mathrm{3}\right)\left(\mathrm{1}+\left({t}+{x}\right)^{\mathrm{2}} \right)}{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\left({t}+\mathrm{3}\right)\left({t}^{\mathrm{2}} \:+\mathrm{2}{xt}\:\:+{x}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${let}\:{decompose}\:{F}\left({t}\right)=\frac{\mathrm{1}}{\left({t}+\mathrm{3}\right)\left({t}^{\mathrm{2}} \:+\mathrm{2}{xt}\:+{x}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${F}\left({t}\right)=\frac{{a}}{{t}+\mathrm{3}}\:+\frac{{bt}\:+{c}}{{t}^{\mathrm{2}} \:+\mathrm{2}{xt}\:+{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:=\frac{\mathrm{1}}{\mathrm{10}−\mathrm{6}{x}\:+{x}^{\mathrm{2}} } \\ $$$${lim}_{{t}\rightarrow+\infty} {tF}\left({t}\right)=\mathrm{0}={a}+{b}\:\Rightarrow{b}=\frac{−\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{6}{x}\:+\mathrm{10}} \\ $$$${F}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{3}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{{a}}{\mathrm{3}}\:+\frac{{c}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\:=\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{3}}{a}\:+{c}\:\Rightarrow \\ $$$$\mathrm{1}=\left({x}^{\mathrm{2}} \:+\mathrm{1}\right){a}\:+\mathrm{3}{c}\:\Rightarrow\mathrm{3}{c}=\mathrm{1}−\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)×\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{6}{x}\:+\mathrm{10}} \\ $$$$\Rightarrow{c}\:=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{6}{x}\:+\mathrm{10}}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{F}\left({t}\right){dt}\:={a}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{{t}+\mathrm{3}}\:+\frac{{b}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{t}\:+\mathrm{2}{x}−\mathrm{2}{x}\:}{{t}^{\mathrm{2}} \:+\mathrm{2}{xt}\:+{x}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\ $$$$\left.={a}\left[{ln}\left({t}+\mathrm{3}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:+\frac{{b}}{\mathrm{2}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{2}{xt}\:+{x}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{{x}} \:+{c}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}{xt}\:+{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}{xt}\:+{x}^{\mathrm{2}} \:+\mathrm{1}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\left({t}+{x}\right)^{\mathrm{2}} \:+\mathrm{1}}\:=_{{t}+{x}={u}} \:\:\:\int_{{x}} ^{{x}+\mathrm{1}} \:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$={arctan}\left({x}+\mathrm{1}\right)−{arctanx}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)={a}\left\{{ln}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)\right\}+\frac{{b}}{\mathrm{2}}\left({ln}\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)−{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+{c}\left({arctan}\left({x}+\mathrm{1}\right)−{arctanx}\right)\right. \\ $$$$\Rightarrow{f}\left({x}\right)={aln}\left(\frac{\mathrm{4}}{\mathrm{3}}\right){x}\:+\frac{{b}}{\mathrm{2}}\int{ln}\left(\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right){dx}\:+{c}\int\:\left({arctan}\left({x}+\mathrm{1}\right)−{arctan}\left({x}\right)\right){dx}\:+{c} \\ $$$$….{be}\:{continued}…. \\ $$
Commented by TawaTawa last updated on 01/Nov/19
God bless you sir. Waiting for the rest sir Thanks for your time sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{Waiting}\:\mathrm{for}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{sir} \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}. \\ $$
Commented by mathmax by abdo last updated on 01/Nov/19
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$
Answered by mind is power last updated on 31/Oct/19
let u=tan^(−1) (x) ⇒x=tg(u)⇒dx=(1+tg^2 (u) )du ∫_2 ^3 ((tan^(−1) (x))/(1−x^2 ))dx=∫_(tan^− (2)) ^(tsn^− (3)) ((u.(1+tg^2 (u))du)/(1−tg^2 (u))) =∫−udu+2∫((udu)/(1−tg^2 (u))) −ln(u)+2∫((udu)/(1−(((e^(iu) −e^(−iu) )/(i(e^(iu) +e^(−iu) ))))^2 ))2∫((udu.)/(((e^(iu) +e^(−iu) )^2 +(e^(iu) −e^(−iu) )^2 )/((e^(iu) +e^(−iu) )^2 ))) =2∫((u(e^(2iu) +e^(−2iu) +2))/(2e^(2iu) +2e^(−2iu) ))du=∫udu+2∫((udu)/(e^(2iu) +e^(−2iu) )) =2∫((ue^(2iu) du)/(1+e^(4iu) ))=2∫((ue^(2iu) du)/((e^(2iu) +i)(e^(2iu) −i)))=∫((udu)/(e^(2iu) +i))+∫((udu)/(e^(2iu) −i)) =∫((udu)/((e^(iu) +e^(−i(π/4)) )(e^(iu) −e^((−iπ)/4) )))+∫((udu)/((e^(iu) +e^(i(π/4)) )(e^(iu) −e^((iπ)/4) ))) =(1/(2e^((−iπ)/4) )){∫((udu)/(e^(iu) −e^((−iπ)/4) ))−∫((udu)/(e^(iu) +e^((−iπ)/4) ))}+(1/(2e^((iπ)/4) )){∫((udu)/(e^(iu) −e^((iπ)/4) ))−∫((udu)/(e^(iu) +e^((iπ)/4) ))} =(e^(i(π/2)) /2){∫(u/(e^(i(u+(π/4))) −1))du}−(e^(−i(π/2)) /2){∫((udu)/(e^(i(u−((3π)/4))) −1))}+(e^(−i(π/2)) /2){∫((udu)/(e^(i(u−(π/4))) −1))}−(e^(i(π/2)) /2)∫((udu)/(e^(i(u+((3π)/4))) −1)) ∫_1 ^(1−z) ((ln(t))/(1−t))=Li_2 (z) ∫_(tan^− (2)) ^(tan^− (3)) ((udu)/(e^(i(u+c)) −1))du w=e^(i(u+c)) ⇒u=−iln(w)−c⇒du=((−idw)/w) ⇒∫_e^(i(tan^− (2)+c)) ^e^(i(tan^− (3)+c)) ((−iln(w)−c)/(w−1)).−i(dw/w) =∫((−ln(w)+ic)/(w(w−1)))=∫((−ln(w)+ic)/(w−1))+∫((ln(w))/w)dw−∫((icdw)/w) =−Li_2 (1−e^(i(tan^− (3)+c)) )+li_2 (1−e^(i(tan^(−2) +c)) )+iclog((e^(i(tan^− (3)+c)) /e^(i(tan^− (2)+c)) ))+log(((itan^− (3)+ic)/(itan^− (2)+ic)))−ic(itan^− (3)−itan^− (2)) put this for c=i(π/4),((−3iπ)/4),((−iπ)/4),((3iπ)/4)
$$\mathrm{let}\:\mathrm{u}=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right) \\ $$$$\Rightarrow\mathrm{x}=\mathrm{tg}\left(\mathrm{u}\right)\Rightarrow\mathrm{dx}=\left(\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\mathrm{u}\right)\:\right)\mathrm{du} \\ $$$$\int_{\mathrm{2}} ^{\mathrm{3}} \frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{tan}^{−} \left(\mathrm{2}\right)} ^{\mathrm{tsn}^{−} \left(\mathrm{3}\right)} \frac{\mathrm{u}.\left(\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\mathrm{u}\right)\right)\mathrm{du}}{\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \left(\mathrm{u}\right)} \\ $$$$=\int−\mathrm{udu}+\mathrm{2}\int\frac{\mathrm{udu}}{\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \left(\mathrm{u}\right)} \\ $$$$−\mathrm{ln}\left(\mathrm{u}\right)+\mathrm{2}\int\frac{\mathrm{udu}}{\mathrm{1}−\left(\frac{\mathrm{e}^{\mathrm{iu}} −\mathrm{e}^{−\mathrm{iu}} }{\mathrm{i}\left(\mathrm{e}^{\mathrm{iu}} +\mathrm{e}^{−\mathrm{iu}} \right)}\right)^{\mathrm{2}} }\mathrm{2}\int\frac{\mathrm{udu}.}{\frac{\left(\mathrm{e}^{\mathrm{iu}} +\mathrm{e}^{−\mathrm{iu}} \right)^{\mathrm{2}} +\left(\mathrm{e}^{\mathrm{iu}} −\mathrm{e}^{−\mathrm{iu}} \right)^{\mathrm{2}} }{\left(\mathrm{e}^{\mathrm{iu}} +\mathrm{e}^{−\mathrm{iu}} \right)^{\mathrm{2}} }} \\ $$$$=\mathrm{2}\int\frac{\mathrm{u}\left(\mathrm{e}^{\mathrm{2iu}} +\mathrm{e}^{−\mathrm{2iu}} +\mathrm{2}\right)}{\mathrm{2e}^{\mathrm{2iu}} +\mathrm{2e}^{−\mathrm{2iu}} }\mathrm{du}=\int\mathrm{udu}+\mathrm{2}\int\frac{\mathrm{udu}}{\mathrm{e}^{\mathrm{2iu}} +\mathrm{e}^{−\mathrm{2iu}} } \\ $$$$=\mathrm{2}\int\frac{\mathrm{ue}^{\mathrm{2iu}} \mathrm{du}}{\mathrm{1}+\mathrm{e}^{\mathrm{4iu}} }=\mathrm{2}\int\frac{\mathrm{ue}^{\mathrm{2iu}} \mathrm{du}}{\left(\mathrm{e}^{\mathrm{2iu}} +\mathrm{i}\right)\left(\mathrm{e}^{\mathrm{2iu}} −\mathrm{i}\right)}=\int\frac{\mathrm{udu}}{\mathrm{e}^{\mathrm{2iu}} +\mathrm{i}}+\int\frac{\mathrm{udu}}{\mathrm{e}^{\mathrm{2iu}} −\mathrm{i}} \\ $$$$=\int\frac{\mathrm{udu}}{\left(\mathrm{e}^{\mathrm{iu}} +\mathrm{e}^{−\mathrm{i}\frac{\pi}{\mathrm{4}}} \right)\left(\mathrm{e}^{\mathrm{iu}} −\mathrm{e}^{\frac{−\mathrm{i}\pi}{\mathrm{4}}} \right)}+\int\frac{\mathrm{udu}}{\left(\mathrm{e}^{\mathrm{iu}} +\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} \right)\left(\mathrm{e}^{\mathrm{iu}} −\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2e}^{\frac{−\mathrm{i}\pi}{\mathrm{4}}} }\left\{\int\frac{\mathrm{udu}}{\mathrm{e}^{\mathrm{iu}} −\mathrm{e}^{\frac{−\mathrm{i}\pi}{\mathrm{4}}} }−\int\frac{\mathrm{udu}}{\mathrm{e}^{\mathrm{iu}} +\mathrm{e}^{\frac{−\mathrm{i}\pi}{\mathrm{4}}} }\right\}+\frac{\mathrm{1}}{\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }\left\{\int\frac{\mathrm{udu}}{\mathrm{e}^{\mathrm{iu}} −\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }−\int\frac{\mathrm{udu}}{\mathrm{e}^{\mathrm{iu}} +\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }\right\} \\ $$$$=\frac{\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} }{\mathrm{2}}\left\{\int\frac{\mathrm{u}}{\mathrm{e}^{\mathrm{i}\left(\mathrm{u}+\frac{\pi}{\mathrm{4}}\right)} −\mathrm{1}}\mathrm{du}\right\}−\frac{\mathrm{e}^{−\mathrm{i}\frac{\pi}{\mathrm{2}}} }{\mathrm{2}}\left\{\int\frac{\mathrm{udu}}{\mathrm{e}^{\mathrm{i}\left(\mathrm{u}−\frac{\mathrm{3}\pi}{\mathrm{4}}\right)} −\mathrm{1}}\right\}+\frac{\mathrm{e}^{−\mathrm{i}\frac{\pi}{\mathrm{2}}} }{\mathrm{2}}\left\{\int\frac{\mathrm{udu}}{\mathrm{e}^{\mathrm{i}\left(\mathrm{u}−\frac{\pi}{\mathrm{4}}\right)} −\mathrm{1}}\right\}−\frac{\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} }{\mathrm{2}}\int\frac{\mathrm{udu}}{\mathrm{e}^{\mathrm{i}\left(\mathrm{u}+\frac{\mathrm{3}\pi}{\mathrm{4}}\right)} −\mathrm{1}} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{1}−\mathrm{z}} \frac{\mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{1}−\mathrm{t}}=\mathrm{Li}_{\mathrm{2}} \left(\mathrm{z}\right) \\ $$$$\int_{\mathrm{tan}^{−} \left(\mathrm{2}\right)} ^{\mathrm{tan}^{−} \left(\mathrm{3}\right)} \frac{\mathrm{udu}}{\mathrm{e}^{\mathrm{i}\left(\mathrm{u}+\mathrm{c}\right)} −\mathrm{1}}\mathrm{du} \\ $$$$\mathrm{w}=\mathrm{e}^{\mathrm{i}\left(\mathrm{u}+\mathrm{c}\right)} \Rightarrow\mathrm{u}=−\mathrm{iln}\left(\mathrm{w}\right)−\mathrm{c}\Rightarrow\mathrm{du}=\frac{−\mathrm{idw}}{\mathrm{w}} \\ $$$$\Rightarrow\int_{\mathrm{e}^{\mathrm{i}\left(\mathrm{tan}^{−} \left(\mathrm{2}\right)+\mathrm{c}\right)} } ^{\mathrm{e}^{\mathrm{i}\left(\mathrm{tan}^{−} \left(\mathrm{3}\right)+\mathrm{c}\right)} } \frac{−\mathrm{iln}\left(\mathrm{w}\right)−\mathrm{c}}{\mathrm{w}−\mathrm{1}}.−\mathrm{i}\frac{\mathrm{dw}}{\mathrm{w}} \\ $$$$=\int\frac{−\mathrm{ln}\left(\mathrm{w}\right)+\mathrm{ic}}{\mathrm{w}\left(\mathrm{w}−\mathrm{1}\right)}=\int\frac{−\mathrm{ln}\left(\mathrm{w}\right)+\mathrm{ic}}{\mathrm{w}−\mathrm{1}}+\int\frac{\mathrm{ln}\left(\mathrm{w}\right)}{\mathrm{w}}\mathrm{dw}−\int\frac{\mathrm{icdw}}{\mathrm{w}} \\ $$$$=−\mathrm{Li}_{\mathrm{2}} \left(\mathrm{1}−\mathrm{e}^{\mathrm{i}\left(\mathrm{tan}^{−} \left(\mathrm{3}\right)+\mathrm{c}\right)} \right)+\mathrm{li}_{\mathrm{2}} \left(\mathrm{1}−\mathrm{e}^{\mathrm{i}\left(\mathrm{tan}^{−\mathrm{2}} +\mathrm{c}\right)} \right)+\mathrm{iclog}\left(\frac{\mathrm{e}^{\mathrm{i}\left(\mathrm{tan}^{−} \left(\mathrm{3}\right)+\mathrm{c}\right)} }{\mathrm{e}^{\mathrm{i}\left(\mathrm{tan}^{−} \left(\mathrm{2}\right)+\mathrm{c}\right)} }\right)+\mathrm{log}\left(\frac{\mathrm{itan}^{−} \left(\mathrm{3}\right)+\mathrm{ic}}{\mathrm{itan}^{−} \left(\mathrm{2}\right)+\mathrm{ic}}\right)−\mathrm{ic}\left(\mathrm{itan}^{−} \left(\mathrm{3}\right)−\mathrm{itan}^{−} \left(\mathrm{2}\right)\right) \\ $$$$\mathrm{put}\:\mathrm{this}\:\mathrm{for}\:\mathrm{c}=\mathrm{i}\frac{\pi}{\mathrm{4}},\frac{−\mathrm{3i}\pi}{\mathrm{4}},\frac{−\mathrm{i}\pi}{\mathrm{4}},\frac{\mathrm{3i}\pi}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by TawaTawa last updated on 31/Oct/19
Wow, God bless you sir. I appreciate your time sir
$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir} \\ $$
Commented by mind is power last updated on 31/Oct/19
most Welcom
$$\mathrm{most}\:\mathrm{Welcom} \\ $$

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