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Question Number 49827 by rahul 19 last updated on 11/Dec/18

The integral ∫_0 ^(1/2) ((ln (1+2x))/(1+4x^2 ))dx = ? a) (π/4)ln2 b)(π/8)ln2 c)(π/(16))ln2 d)(π/(32))ln2

$${The}\:{integral}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{2}{x}\right)}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }{dx}\:=\:? \\ $$$$\left.{a}\left.\right)\left.\:\left.\frac{\pi}{\mathrm{4}}{ln}\mathrm{2}\:\:\:\:{b}\right)\frac{\pi}{\mathrm{8}}{ln}\mathrm{2}\:\:\:\:{c}\right)\frac{\pi}{\mathrm{16}}{ln}\mathrm{2}\:\:\:{d}\right)\frac{\pi}{\mathrm{32}}{ln}\mathrm{2} \\ $$

Commented by rahul 19 last updated on 11/Dec/18

thank you sir!

$${thank}\:{you}\:{sir}! \\ $$

Commented by rahul 19 last updated on 11/Dec/18

Any short/tricky method other than usual substitution:2x=tanθ ???

$${Any}\:{short}/{tricky}\:{method}\:{other} \\ $$$${than}\:{usual}\:{substitution}:\mathrm{2}{x}=\mathrm{tan}\theta\:??? \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18

remember the answer...

$${remember}\:{the}\:{answer}... \\ $$

Commented by Abdo msup. last updated on 11/Dec/18

let I =∫_0 ^(1/2) ((ln(1+2x))/(1+4x^2 ))dx changement 2x =tant give I = (1/2)∫_0 ^(π/4) ((ln(1+tant))/(1+tan^2 t)) (1+tan^2 t)dt ⇒ 2I = ∫_0 ^(π/4) ln(1+tant )dt =∫_0 ^(π/4) ln(((cost +sint)/(cost)))dt =∫_0 ^(π/4) ln(cost +sint)dt−∫_0 ^(π/4) ln(cost)dt =∫_0 ^(π/4) ln((√2)sin(t+(π/4))dt −∫_0 ^(π/4) ln(cost)dt =(π/8)ln(2) +∫_0 ^(π/4) ln(sin(t+(π/4)))dt −∫_0 ^(π/4) ln(cost)dt ∫_0 ^(π/4) ln(sin(t+(π/4)))dt =_(t+(π/4)=u) ∫_(π/4) ^(π/2) ln(sinu)du =_(u =(π/2)−α) ∫_(π/4) ^0 ln(cosα)(−dα)=∫_0 ^(π/4) ln(cosα)dα 2I =(π/8)ln(2) ⇒ I =(π/(16))ln(2) .

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }{dx}\:{changement}\:\mathrm{2}{x}\:={tant}\:{give} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{ln}\left(\mathrm{1}+{tant}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left(\mathrm{1}+{tant}\:\right){dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\frac{{cost}\:+{sint}}{{cost}}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({cost}\:+{sint}\right){dt}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cost}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\sqrt{\mathrm{2}}{sin}\left({t}+\frac{\pi}{\mathrm{4}}\right){dt}\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cost}\right){dt}\right. \\ $$$$=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({sin}\left({t}+\frac{\pi}{\mathrm{4}}\right)\right){dt}\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cost}\right){dt} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({sin}\left({t}+\frac{\pi}{\mathrm{4}}\right)\right){dt}\:=_{{t}+\frac{\pi}{\mathrm{4}}={u}} \:\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinu}\right){du} \\ $$$$=_{{u}\:=\frac{\pi}{\mathrm{2}}−\alpha} \:\:\:\:\int_{\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} {ln}\left({cos}\alpha\right)\left(−{d}\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({cos}\alpha\right){d}\alpha \\ $$$$\mathrm{2}{I}\:=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{16}}{ln}\left(\mathrm{2}\right)\:. \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18

t=2x dt=2dx ∫_0 ^1 ((ln(1+t))/(1+t^2 ))dt I=∫_0 ^1 ((ln(1+t))/(1+t^2 ))dt t=tana dt=sec^2 ada I=∫_0 ^(π/4) ((ln(1+tana))/(sec^2 a))×sec^2 ada =∫_0 ^(π/4) ln{1+tan((π/4)−a)} =∫_0 ^(π/4) ln{1+((1−tana)/(1+tana))}da =∫_0 ^(π/4) ln((2/(1+tana)))da =∫_0 ^(π/4) ln2 da−∫_0 ^(π/4) ln(1+tana)da 2I=ln2∫_0 ^(π/4) da I=(1/2)×(ln2)×(π/4)=(π/8)ln2

$${t}=\mathrm{2}{x}\:\:{dt}=\mathrm{2}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$${t}={tana}\:\:{dt}={sec}^{\mathrm{2}} {ada} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left(\mathrm{1}+{tana}\right)}{{sec}^{\mathrm{2}} {a}}×{sec}^{\mathrm{2}} {ada} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left\{\mathrm{1}+{tan}\left(\frac{\pi}{\mathrm{4}}−{a}\right)\right\} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left\{\mathrm{1}+\frac{\mathrm{1}−{tana}}{\mathrm{1}+{tana}}\right\}{da} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\frac{\mathrm{2}}{\mathrm{1}+{tana}}\right){da} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\mathrm{2}\:{da}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tana}\right){da} \\ $$$$\mathrm{2}{I}={ln}\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {da} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}×\left({ln}\mathrm{2}\right)×\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{8}}{ln}\mathrm{2} \\ $$$$ \\ $$

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