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Question Number 98311 by mathmax by abdo last updated on 12/Jun/20
calculate ∫_0 ^∞ ((lnx)/((1+x)^2 ))dx
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx}\: \\ $$
Answered by mathmax by abdo last updated on 13/Jun/20
A =∫_0 ^∞ ((ln(x))/((1+x)^2 ))dx ⇒ A =∫_0 ^1 ((ln(x))/((1+x)^2 ))dx +∫_1 ^∞ ((ln(x))/((1+x)^2 ))dx(→x=(1/t)) =∫_0 ^1 ((ln(x))/((1+x)^2 ))dx +∫_0 ^1 ((−ln(t))/((1+(1/t))^2 ))(dt/t^2 ) =∫_0 ^1 ((ln(x))/((1+x)^2 ))dx−∫_0 ^1 ((ln(t))/((t+1)^2 ))dt =0 ⇒ A =0
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow\:\mathrm{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx}\:+\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx}\left(\rightarrow\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{−\mathrm{ln}\left(\mathrm{t}\right)}{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{2}} }\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{t}\right)}{\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt}\:=\mathrm{0}\:\Rightarrow\:\mathrm{A}\:=\mathrm{0} \\ $$

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