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Question Number 163536 by HongKing last updated on 07/Jan/22

Prove that: ∫_( 9) ^( 1) ((ln (1 + x))/(1 + x^2 )) dx = ((π ∙ ln (2))/8)

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\:\mathrm{9}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}\:+\:\mathrm{x}\right)}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:\frac{\pi\:\centerdot\:\mathrm{ln}\:\left(\mathrm{2}\right)}{\mathrm{8}} \\ $$

Answered by Ar Brandon last updated on 07/Jan/22

I=∫_0 ^1 ((ln(1+x))/(1+x^2 ))dx , x=tanϑ =∫_0 ^(π/4) ln(1+tanϑ)dϑ =∫_0 ^(π/4) ln(cosϑ+sinϑ)dϑ−∫_0 ^(π/4) ln(cosϑ)dϑ =∫_0 ^(π/4) ln((√2)cos(x−(π/4)))dx−∫_0 ^(π/4) ln(cosϑ)dϑ =((πln2)/8)+∫_0 ^(π/4) ln(cos(x−(π/4)))−∫_0 ^(π/4) ln(cosϑ)dϑ =((πln2)/8)+∫_0 ^(π/4) ln(cosϑ)dϑ−∫_0 ^(π/4) ln(cosϑ)dϑ=((πln2)/8)

$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:,\:{x}=\mathrm{tan}\vartheta \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{tan}\vartheta\right){d}\vartheta \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}\vartheta+\mathrm{sin}\vartheta\right){d}\vartheta−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}\vartheta\right){d}\vartheta \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\sqrt{\mathrm{2}}\mathrm{cos}\left({x}−\frac{\pi}{\mathrm{4}}\right)\right){dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}\vartheta\right){d}\vartheta \\ $$$$\:\:\:=\frac{\pi\mathrm{ln2}}{\mathrm{8}}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}\left({x}−\frac{\pi}{\mathrm{4}}\right)\right)−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}\vartheta\right){d}\vartheta \\ $$$$\:\:\:=\frac{\pi\mathrm{ln2}}{\mathrm{8}}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}\vartheta\right){d}\vartheta−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}\vartheta\right){d}\vartheta=\frac{\pi\mathrm{ln2}}{\mathrm{8}} \\ $$

Commented by HongKing last updated on 07/Jan/22

perfect my dear Sir thank you so much

$$\mathrm{perfect}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

Commented by peter frank last updated on 11/Jan/22

great

$$\mathrm{great} \\ $$

Answered by smallEinstein last updated on 07/Jan/22

Commented by GalaxyBills last updated on 07/Jan/22

My Boss that

$${My}\:{Boss}\:{that} \\ $$

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