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Question Number 33328 by prof Abdo imad last updated on 14/Apr/18
find   ∫_(π/4) ^(4/π)   (1+(1/x^2 ))arctanx dx
$${find}\:\:\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctanx}\:{dx} \\ $$
Commented by math khazana by abdo last updated on 19/Apr/18
let put I = ∫_(π/4) ^(4/π)   (1+(1/x^2 ))arctanx dx .let integrate by  parts u^′  =1+(1/x^2 )  and v =arctanx  I = [(1−(1/x))arctanx]_(π/4) ^(4/π)   −∫_(π/4) ^(4/π)  (1−(1/x)) (dx/(1+x^2 ))  = (1−(π/4))arctan((4/π)) −(1−(4/π)) −∫_(π/4) ^(4/π)   (dx/(1+x^2 ))  + ∫_(π/4) ^(4/π)       (dx/(x( 1+x^2 )))  but  ∫_(π/4) ^(4/π)    (dx/(1+x^2 )) = arctan( (4/π)) −arctan((π/4))  =(π/2) −1−1=(π/2) −2  let?decompose  F(x) =  (1/(x(1+x^2 ))) = (a/x)  +((bx +c)/(1+x^2 ))  a =lim_(x→0) x F(x) = 1  lim_(x→+∞) x F(x) =0 = a +b ⇒b=−a =−1  F(x) = (1/x)  +((−x +c)/(1+x^2 ))    we look tbat c=0 ⇒  F(x) = (1/x)  −(x/(1+x^2 )) ⇒  ∫_(π/4) ^(4/π)     (dx/(x(1+x^2 ))) = ∫_(π/4) ^(4/π)  ((1/x) −(x/(1+x^2 )))dx  =[ ln(x)−(1/2)ln(1+x^2 )]_(π/4) ^(4/π)  =[ln((x/( (√(1+x^2 )))))]_(π/4) ^(4/π)   = ln(  (4/(π(√(1+((16)/π^2 )))))) −ln(   (π/(4(√(1+(π^2 /(16)))))))  I =(1−(π/4))((π/2) −1) +1 +(4/π) −(π/2)   +ln( (4/(π(√(1+((16)/π^2 )))))) −ln(  (π/(4(√(1+(π^2 /(16))))))) .
$${let}\:{put}\:{I}\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctanx}\:{dx}\:.{let}\:{integrate}\:{by} \\ $$$${parts}\:{u}^{'} \:=\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\:{and}\:{v}\:={arctanx} \\ $$$${I}\:=\:\left[\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right){arctanx}\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\:−\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\:\left(\mathrm{1}−\frac{\pi}{\mathrm{4}}\right){arctan}\left(\frac{\mathrm{4}}{\pi}\right)\:−\left(\mathrm{1}−\frac{\mathrm{4}}{\pi}\right)\:−\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$+\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\:\:\:\:\:\frac{{dx}}{{x}\left(\:\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\:{but} \\ $$$$\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:{arctan}\left(\:\frac{\mathrm{4}}{\pi}\right)\:−{arctan}\left(\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\:−\mathrm{1}−\mathrm{1}=\frac{\pi}{\mathrm{2}}\:−\mathrm{2}\:\:{let}?{decompose} \\ $$$${F}\left({x}\right)\:=\:\:\frac{\mathrm{1}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:=\:\frac{{a}}{{x}}\:\:+\frac{{bx}\:+{c}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${a}\:={lim}_{{x}\rightarrow\mathrm{0}} {x}\:{F}\left({x}\right)\:=\:\mathrm{1} \\ $$$${lim}_{{x}\rightarrow+\infty} {x}\:{F}\left({x}\right)\:=\mathrm{0}\:=\:{a}\:+{b}\:\Rightarrow{b}=−{a}\:=−\mathrm{1} \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}}\:\:+\frac{−{x}\:+{c}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:\:\:{we}\:{look}\:{tbat}\:{c}=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}}\:\:−\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\:\:\:\frac{{dx}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\left(\frac{\mathrm{1}}{{x}}\:−\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx} \\ $$$$=\left[\:{ln}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:=\left[{ln}\left(\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \\ $$$$=\:{ln}\left(\:\:\frac{\mathrm{4}}{\pi\sqrt{\mathrm{1}+\frac{\mathrm{16}}{\pi^{\mathrm{2}} }}}\right)\:−{ln}\left(\:\:\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{16}}}}\right) \\ $$$${I}\:=\left(\mathrm{1}−\frac{\pi}{\mathrm{4}}\right)\left(\frac{\pi}{\mathrm{2}}\:−\mathrm{1}\right)\:+\mathrm{1}\:+\frac{\mathrm{4}}{\pi}\:−\frac{\pi}{\mathrm{2}}\: \\ $$$$+{ln}\left(\:\frac{\mathrm{4}}{\pi\sqrt{\mathrm{1}+\frac{\mathrm{16}}{\pi^{\mathrm{2}} }}}\right)\:−{ln}\left(\:\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{16}}}}\right)\:. \\ $$

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