Question Number 52459 by Abdo msup. last updated on 08/Jan/19
$${let}\:{f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{i}\alpha{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{1}\right){determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left(\alpha\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{ix}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:{and}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+\mathrm{2}{ix}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}. \\ $$
Commented by maxmathsup by imad last updated on 08/Jan/19
![1) we have f^′ (α) =∫_0 ^1 ((ix)/((1+iαx)(1+x^2 )))dx so for α≠0 f^′ (α)=(1/α) ∫_0 ^1 ((1+iαx−1)/((1+iαx)(x^2 +1)))dx =(1/α) −(1/α) ∫_0 ^1 (dx/((iαx+1)(x^2 +1)))dx let decompose F(x)=(1/((iαx+1)(x^2 +1))) ⇒ F(x)=(a/(iαx+1)) +((bx+c)/(x^2 +1)) a=lim_(x→((−1)/(iα))) (iαx+1)F(x) = (1/((((1/(iα )))^2 +1))) =(1/((1/(−α^2 ))+1)) =−(1/(1−α^2 )) α^2 =(α^2 /(α^2 −1)) lim_(x→+∞ ) xF(x)=0 =(a/(iα)) +b ⇒b=−(a/(iα)) =((ia)/α) ⇒ F(x)=(α^2 /(α^2 −1)) (1/(iαx+1)) +((((ia)/α)x+c)/(x^2 +1)) (we suppose α≠+^− 1) F(o)=1 =(α^2 /(α^2 −1)) +c ⇒c=1−(α^2 /(α^2 −1)) =((−1)/(α^2 −1)) ⇒ F(x) =(α^2 /((α^2 −1)(1+iαx))) +(((i/α)(α^2 /(α^2 −1))x−(1/(α^2 −1)))/(x^2 +1)) =(α^2 /((α^2 −1)(1+iαx))) +(1/(α^2 −1)) ((iαx−1)/(x^2 +1)) ⇒ f^′ (α)=(1/α) −(1/α) ∫_0 ^1 (α^2 /((α^2 −1)(1+iαx))) dx−(1/(α(α^2 −1))) ∫_0 ^1 ((iαx −1)/(x^2 +1))dx =(1/α) −(α/(α^2 −1))(1/(iα)) ∫_0 ^1 ((iαdx)/(1+iαx)) dx −(i/(2(α^2 −1)))[ln(x^2 +1)]_0 ^1 +(π/(4α(α^2 −1))) =(1/α) +(i/(α^2 −1))ln(1+iα) −((iln(2))/(2(α^2 −1))) +(π/(4α(α^2 −1))) ⇒ f(α) =ln∣α∣ +i ∫ ((ln(1+iα))/(α^2 −1))dα +i((ln(2))/2) ∫ (dα/(α^2 −1)) +(π/4) ∫ (dα/(α(α^2 −1))) +c but ∫ (dα/(α^2 −1)) =(1/2) ∫ ((1/(α−1)) −(1/(α+1)))dα =(1/2)ln∣((α−1)/(α+1))∣ let decompose w(α)=(1/(α(α^2 −1))) =(1/(α(α−1)(α+1))) w(α)=(a/α) +(b/(α−1)) +(c/(α+1)) ⇒a =−1 b=(1/2) and c=(1/2) ⇒w(α)=−(1/α) +(1/(2(α−1))) +(1/(2(α+1))) ⇒ ∫ (dα/(α(α^2 −1))) =−ln∣α∣ +(1/2)ln∣α^2 −1∣ ⇒ f(α)=i ∫ ((ln(1+iα))/(α^2 −1))dα +((iln(2))/4)ln∣((α−1)/(α+1))∣+ (1−(π/4))ln∣α∣+(π/8)ln∣α^2 −1∣ +c](https://www.tinkutara.com/question/Q52483.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left(\alpha\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ix}}{\left(\mathrm{1}+{i}\alpha{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\:\:{so}\:{for}\:\alpha\neq\mathrm{0}\: \\ $$$${f}^{'} \left(\alpha\right)=\frac{\mathrm{1}}{\alpha}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}+{i}\alpha{x}−\mathrm{1}}{\left(\mathrm{1}+{i}\alpha{x}\right)\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}{dx}\:=\frac{\mathrm{1}}{\alpha}\:−\frac{\mathrm{1}}{\alpha}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\left({i}\alpha{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}{dx}\:{let} \\ $$$${decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{\left({i}\alpha{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{i}\alpha{x}+\mathrm{1}}\:+\frac{{bx}+{c}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}={lim}_{{x}\rightarrow\frac{−\mathrm{1}}{{i}\alpha}} \:\:\:\left({i}\alpha{x}+\mathrm{1}\right){F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\left(\left(\frac{\mathrm{1}}{{i}\alpha\:}\right)^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\frac{\mathrm{1}}{−\alpha^{\mathrm{2}} }+\mathrm{1}}\:=−\frac{\mathrm{1}}{\mathrm{1}−\alpha^{\mathrm{2}} }\:\alpha^{\mathrm{2}} \:=\frac{\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} −\mathrm{1}} \\ $$$${lim}_{{x}\rightarrow+\infty\:} \:\:{xF}\left({x}\right)=\mathrm{0}\:=\frac{{a}}{{i}\alpha}\:+{b}\:\Rightarrow{b}=−\frac{{a}}{{i}\alpha}\:=\frac{{ia}}{\alpha}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} −\mathrm{1}}\:\frac{\mathrm{1}}{{i}\alpha{x}+\mathrm{1}}\:+\frac{\frac{{ia}}{\alpha}{x}+{c}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\:\left({we}\:{suppose}\:\alpha\neq\overset{−} {+}\mathrm{1}\right) \\ $$$${F}\left({o}\right)=\mathrm{1}\:=\frac{\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} −\mathrm{1}}\:+{c}\:\Rightarrow{c}=\mathrm{1}−\frac{\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} −\mathrm{1}}\:=\frac{−\mathrm{1}}{\alpha^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{\alpha^{\mathrm{2}} }{\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{1}+{i}\alpha{x}\right)}\:+\frac{\frac{{i}}{\alpha}\frac{\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} −\mathrm{1}}{x}−\frac{\mathrm{1}}{\alpha^{\mathrm{2}} −\mathrm{1}}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\frac{\alpha^{\mathrm{2}} }{\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{1}+{i}\alpha{x}\right)}\:\:+\frac{\mathrm{1}}{\alpha^{\mathrm{2}} −\mathrm{1}}\:\frac{{i}\alpha{x}−\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${f}^{'} \left(\alpha\right)=\frac{\mathrm{1}}{\alpha}\:−\frac{\mathrm{1}}{\alpha}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\alpha^{\mathrm{2}} }{\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{1}+{i}\alpha{x}\right)}\:{dx}−\frac{\mathrm{1}}{\alpha\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{i}\alpha{x}\:−\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\alpha}\:−\frac{\alpha}{\alpha^{\mathrm{2}} −\mathrm{1}}\frac{\mathrm{1}}{{i}\alpha}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{i}\alpha{dx}}{\mathrm{1}+{i}\alpha{x}}\:{dx}\:−\frac{{i}}{\mathrm{2}\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\left[{ln}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:+\frac{\pi}{\mathrm{4}\alpha\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\alpha}\:+\frac{{i}}{\alpha^{\mathrm{2}} −\mathrm{1}}{ln}\left(\mathrm{1}+{i}\alpha\right)\:−\frac{{iln}\left(\mathrm{2}\right)}{\mathrm{2}\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\:+\frac{\pi}{\mathrm{4}\alpha\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\:\Rightarrow \\ $$$${f}\left(\alpha\right)\:={ln}\mid\alpha\mid\:+{i}\:\int\:\:\frac{{ln}\left(\mathrm{1}+{i}\alpha\right)}{\alpha^{\mathrm{2}} −\mathrm{1}}{d}\alpha\:+{i}\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:\int\:\frac{{d}\alpha}{\alpha^{\mathrm{2}} −\mathrm{1}}\:+\frac{\pi}{\mathrm{4}}\:\int\:\:\:\frac{{d}\alpha}{\alpha\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\:+{c}\:{but} \\ $$$$\int\:\:\:\frac{{d}\alpha}{\alpha^{\mathrm{2}} −\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\left(\frac{\mathrm{1}}{\alpha−\mathrm{1}}\:−\frac{\mathrm{1}}{\alpha+\mathrm{1}}\right){d}\alpha\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\alpha−\mathrm{1}}{\alpha+\mathrm{1}}\mid\:{let}\:{decompose} \\ $$$${w}\left(\alpha\right)=\frac{\mathrm{1}}{\alpha\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\alpha\left(\alpha−\mathrm{1}\right)\left(\alpha+\mathrm{1}\right)} \\ $$$${w}\left(\alpha\right)=\frac{{a}}{\alpha}\:+\frac{{b}}{\alpha−\mathrm{1}}\:+\frac{{c}}{\alpha+\mathrm{1}}\:\Rightarrow{a}\:=−\mathrm{1} \\ $$$${b}=\frac{\mathrm{1}}{\mathrm{2}}\:\:{and}\:\:{c}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{w}\left(\alpha\right)=−\frac{\mathrm{1}}{\alpha}\:+\frac{\mathrm{1}}{\mathrm{2}\left(\alpha−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}\left(\alpha+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\int\:\frac{{d}\alpha}{\alpha\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\:=−{ln}\mid\alpha\mid\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\alpha^{\mathrm{2}} −\mathrm{1}\mid\:\Rightarrow \\ $$$${f}\left(\alpha\right)={i}\:\int\:\:\frac{{ln}\left(\mathrm{1}+{i}\alpha\right)}{\alpha^{\mathrm{2}} −\mathrm{1}}{d}\alpha\:+\frac{{iln}\left(\mathrm{2}\right)}{\mathrm{4}}{ln}\mid\frac{\alpha−\mathrm{1}}{\alpha+\mathrm{1}}\mid+\:\left(\mathrm{1}−\frac{\pi}{\mathrm{4}}\right){ln}\mid\alpha\mid+\frac{\pi}{\mathrm{8}}{ln}\mid\alpha^{\mathrm{2}} −\mathrm{1}\mid\:+{c} \\ $$$$ \\ $$
Commented by Abdo msup. last updated on 08/Jan/19
![2) we have 1+ix=(√(1+x^2 ))((1/( (√(1+x^2 )))) +i(x/( (√(1+x^2 )))))=r e^(iθ) ⇒ r=(√(1+x^2 )) and θ=arctanx ⇒ ln(1+ix)=(1/2)ln(1+x^2 )+iarctanx ⇒ ∫_0 ^1 ((ln(1+ix))/(1+x^2 )) =(1/2) ∫_0 ^1 ((ln(1+x^2 ))/(1+x^2 )) dx +i∫_0 ^1 ((arctanx)/(1+x^2 ))dx by parts I=∫_0 ^1 ((arctanx)/(1+x^2 ))dx=[arctan^2 x]_0 ^1 −∫_0 ^1 ((arctanx)/(1+x^2 )) =(π^2 /(16)) −I ⇒2I=(π^2 /(16)) ⇒I=(π^2 /(32)) let find ∫_0 ^1 ((ln(1+x^2 ))/(1+x^2 )) dx let f(t)=∫_0 ^1 ((ln(1+tx^2 ))/(1+x^2 )) dx with t>0 we have f^′ (t) = ∫_0 ^1 (x^2 /((1+tx^2 )(1+x^2 )))dx =_((√t)x=u) ∫_0 ^(√t) (u^2 /(t(1+u^2 )(1+(u^2 /t)))) (du/( (√t))) =(1/( (√t))) ∫_0 ^(√t) (u^2 /((u^2 +1)(u^2 +t))) du let decompose F(u) =(u^2 /((u^2 +1)(u^2 +t))) ⇒ F(u)=((au +b)/(u^2 +1)) +((cu+d)/(u^2 +t)) F(−u)=F(u) ⇒((−au+b)/(u^2 +1)) +((−cu +d)/(u^2 +t)) =F(u) ⇒a=0 and c=0 ⇒ F(u)=(b/(u^2 +1)) +(d/(u^2 +t)) we see that F(u)=(1/(t−1))((1/(u^2 +1))−(1/(u^2 +t))) ⇒ ∫_0 ^(√t) F(u)du =(1/(t−1)) ∫_0 ^(√t) (du/(u^2 +1)) −(1/(t−1)) ∫_0 ^(√t) (du/(u^2 +t)) but ∫_0 ^(√t) (du/(1+u^2 )) =arctan((√t)) ∫_0 ^(√t) (du/(u^2 +t)) =_(u=(√t)α) ∫_0 ^1 (((√t)dα)/(t(1+α^2 ))) =(1/( (√t)))(π/4) ⇒ f^′ (t)=(1/( (√t))){((arctan((√t)))/(t−1)) −(π/(4(√t)(t−1)))} ⇒ f(t) =∫ ((arctan((√t)))/( (√t)(t−1)))dy −(π/4) ∫ (dt/(t(t−1))) +C ∫ (dt/(t(t−1))) =∫ ((1/(t−1)) −(1/t))dt=ln∣((t−1)/t)∣ +c_1 ∫ ((arctan((√t)))/( (√t)(t−1)))dt =_((√t)=u) ∫ ((arctan(u))/(u(u^2 −1))) (2u)du =2 ∫ ((arctan(u))/(u^2 −1)) du ....be continued...](https://www.tinkutara.com/question/Q52497.png)
$$\left.\mathrm{2}\right)\:{we}\:{have}\:\mathrm{1}+{ix}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+{i}\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)={r}\:{e}^{{i}\theta} \:\Rightarrow \\ $$$${r}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{and}\:\theta={arctanx}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{ix}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+{iarctanx}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}+{ix}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:+{i}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctanx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${by}\:{parts}\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctanx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\left[{arctan}^{\mathrm{2}} {x}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{arctanx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\:−{I}\:\Rightarrow\mathrm{2}{I}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\:\Rightarrow{I}=\frac{\pi^{\mathrm{2}} }{\mathrm{32}} \\ $$$${let}\:{find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:{dx}\:{let}\: \\ $$$${f}\left({t}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}+{tx}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:{with}\:{t}>\mathrm{0}\:{we}\:{have} \\ $$$${f}^{'} \left({t}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{tx}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$=_{\sqrt{{t}}{x}={u}} \:\:\:\:\:\int_{\mathrm{0}} ^{\sqrt{{t}}} \:\:\:\:\:\:\frac{{u}^{\mathrm{2}} }{{t}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{{u}^{\mathrm{2}} }{{t}}\right)}\:\frac{{du}}{\:\sqrt{{t}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{t}}}\:\int_{\mathrm{0}} ^{\sqrt{{t}}} \:\:\:\:\:\frac{{u}^{\mathrm{2}} }{\left({u}^{\mathrm{2}} +\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+{t}\right)}\:{du}\:\:{let}\:{decompose} \\ $$$${F}\left({u}\right)\:=\frac{{u}^{\mathrm{2}} }{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+{t}\right)}\:\Rightarrow \\ $$$${F}\left({u}\right)=\frac{{au}\:+{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{cu}+{d}}{{u}^{\mathrm{2}} \:+{t}} \\ $$$${F}\left(−{u}\right)={F}\left({u}\right)\:\Rightarrow\frac{−{au}+{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{cu}\:+{d}}{{u}^{\mathrm{2}} \:+{t}} \\ $$$$={F}\left({u}\right)\:\Rightarrow{a}=\mathrm{0}\:{and}\:{c}=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({u}\right)=\frac{{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{d}}{{u}^{\mathrm{2}} \:+{t}}\:\:{we}\:{see}\:{that}\: \\ $$$${F}\left({u}\right)=\frac{\mathrm{1}}{{t}−\mathrm{1}}\left(\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+\mathrm{1}}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+{t}}\right)\:\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\sqrt{{t}}} \:{F}\left({u}\right){du}\:=\frac{\mathrm{1}}{{t}−\mathrm{1}}\:\int_{\mathrm{0}} ^{\sqrt{{t}}} \:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{\mathrm{1}}{{t}−\mathrm{1}}\:\int_{\mathrm{0}} ^{\sqrt{{t}}} \:\:\frac{{du}}{{u}^{\mathrm{2}} \:+{t}} \\ $$$${but}\:\int_{\mathrm{0}} ^{\sqrt{{t}}} \:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:={arctan}\left(\sqrt{{t}}\right) \\ $$$$\int_{\mathrm{0}} ^{\sqrt{{t}}} \:\:\frac{{du}}{{u}^{\mathrm{2}} \:+{t}}\:=_{{u}=\sqrt{{t}}\alpha} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\sqrt{{t}}{d}\alpha}{{t}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\:\sqrt{{t}}}\frac{\pi}{\mathrm{4}}\:\Rightarrow \\ $$$${f}^{'} \left({t}\right)=\frac{\mathrm{1}}{\:\sqrt{{t}}}\left\{\frac{{arctan}\left(\sqrt{{t}}\right)}{{t}−\mathrm{1}}\:−\frac{\pi}{\mathrm{4}\sqrt{{t}}\left({t}−\mathrm{1}\right)}\right\}\:\Rightarrow \\ $$$${f}\left({t}\right)\:=\int\:\:\:\frac{{arctan}\left(\sqrt{{t}}\right)}{\:\sqrt{{t}}\left({t}−\mathrm{1}\right)}{dy}\:−\frac{\pi}{\mathrm{4}}\:\int\:\:\frac{{dt}}{{t}\left({t}−\mathrm{1}\right)}\:+{C} \\ $$$$\int\:\:\:\frac{{dt}}{{t}\left({t}−\mathrm{1}\right)}\:=\int\:\left(\frac{\mathrm{1}}{{t}−\mathrm{1}}\:−\frac{\mathrm{1}}{{t}}\right){dt}={ln}\mid\frac{{t}−\mathrm{1}}{{t}}\mid\:+{c}_{\mathrm{1}} \\ $$$$\int\:\:\frac{{arctan}\left(\sqrt{{t}}\right)}{\:\sqrt{{t}}\left({t}−\mathrm{1}\right)}{dt}\:=_{\sqrt{{t}}={u}} \:\:\:\int\:\:\:\:\frac{{arctan}\left({u}\right)}{{u}\left({u}^{\mathrm{2}} −\mathrm{1}\right)}\:\left(\mathrm{2}{u}\right){du} \\ $$$$=\mathrm{2}\:\int\:\:\frac{{arctan}\left({u}\right)}{{u}^{\mathrm{2}} −\mathrm{1}}\:{du}\:….{be}\:{continued}… \\ $$$$ \\ $$$$ \\ $$$$ \\ $$