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Question Number 3017 by Filup last updated on 03/Dec/15

A=∫_μ ^( μ+ε) (dx/(x+n)) A=ln(x+n) ∣_μ ^(μ+ε) =ln(μ+ε+n)−ln(μ+n) =ln(((μ+ε+n)/(μ+n))) =ln(((μ+n)/(μ+n))+(ε/(μ+n))) ∴A=∫_μ ^( μ+ε) (dx/(x+n))=ln((ε/(μ+n))+1) For (a/b)=c i) if a>b, c>1 ii) if a<b, c<1 iii) if a=b, c=1 Case (i) ε>μ+n (ε/(μ+n))>1 ∴A>ln(2) Case (ii) ε<μ+n (ε/(μ+n))<1 ∴A<ln(2) Case (iii) ε=μ+n (ε/(μ+n))=1 A=ln(2) What other observations can you find regarding similar logarithmic functions?

$${A}=\int_{\mu} ^{\:\mu+\epsilon} \:\frac{{dx}}{{x}+{n}} \\ $$ $$ \\ $$ $${A}=\mathrm{ln}\left({x}+{n}\right)\:\mid_{\mu} ^{\mu+\epsilon} \\ $$ $$=\mathrm{ln}\left(\mu+\epsilon+{n}\right)−\mathrm{ln}\left(\mu+{n}\right) \\ $$ $$=\mathrm{ln}\left(\frac{\mu+\epsilon+{n}}{\mu+{n}}\right) \\ $$ $$=\mathrm{ln}\left(\frac{\mu+{n}}{\mu+{n}}+\frac{\epsilon}{\mu+{n}}\right) \\ $$ $$ \\ $$ $$\therefore{A}=\int_{\mu} ^{\:\mu+\epsilon} \:\frac{{dx}}{{x}+{n}}=\mathrm{ln}\left(\frac{\epsilon}{\mu+{n}}+\mathrm{1}\right) \\ $$ $$ \\ $$ $$\mathrm{For}\:\:\frac{{a}}{{b}}={c} \\ $$ $$\left.{i}\right)\:\mathrm{if}\:\:{a}>{b},\:{c}>\mathrm{1} \\ $$ $$\left.{ii}\right)\:\mathrm{if}\:\:\:{a}<{b},\:{c}<\mathrm{1} \\ $$ $$\left.{iii}\right)\:\mathrm{if}\:\:{a}={b},\:{c}=\mathrm{1} \\ $$ $$ \\ $$ $$\mathrm{Case}\:\left({i}\right) \\ $$ $$\epsilon>\mu+{n} \\ $$ $$\frac{\epsilon}{\mu+{n}}>\mathrm{1} \\ $$ $$\therefore{A}>\mathrm{ln}\left(\mathrm{2}\right) \\ $$ $$ \\ $$ $$\mathrm{Case}\:\left({ii}\right) \\ $$ $$\epsilon<\mu+{n} \\ $$ $$\frac{\epsilon}{\mu+{n}}<\mathrm{1} \\ $$ $$\therefore{A}<\mathrm{ln}\left(\mathrm{2}\right) \\ $$ $$ \\ $$ $$\mathrm{Case}\:\left({iii}\right) \\ $$ $$\epsilon=\mu+{n} \\ $$ $$\frac{\epsilon}{\mu+{n}}=\mathrm{1} \\ $$ $${A}=\mathrm{ln}\left(\mathrm{2}\right) \\ $$ $$ \\ $$ $$\mathrm{What}\:\mathrm{other}\:\mathrm{observations}\:\mathrm{can}\:\mathrm{you}\:\mathrm{find} \\ $$ $$\mathrm{regarding}\:\mathrm{similar}\:\mathrm{logarithmic}\:\mathrm{functions}? \\ $$

Commented byFilup last updated on 03/Dec/15

Maybe look at cases of ∫_μ ^( με) , ∫_μ ^( μ^ε ) , etc.

$$\mathrm{Maybe}\:\mathrm{look}\:\mathrm{at}\:\mathrm{cases}\:\mathrm{of}\:\int_{\mu} ^{\:\mu\epsilon} ,\:\int_{\mu} ^{\:\mu^{\epsilon} } ,\:{etc}. \\ $$

Commented by123456 last updated on 03/Dec/15

μ>−n∨μ+ε<−n

$$\mu>−{n}\vee\mu+\epsilon<−{n} \\ $$

Commented byFilup last updated on 03/Dec/15

Please elaborate?

$$\mathrm{Please}\:\mathrm{elaborate}? \\ $$

Answered by Filup last updated on 03/Dec/15

∫_μ ^(με) (dx/(x+n))=ln(((με+n)/(μ+n))) =ln(((ε+(n/μ))/(1+(n/μ)))) k=(n/μ) A=ln(((ε+k)/(1+k))) Similar to original post: ε>1, A>0 ε<1, A<0 ε=1, A=0

$$\int_{\mu} ^{\mu\epsilon} \:\frac{{dx}}{{x}+{n}}=\mathrm{ln}\left(\frac{\mu\epsilon+{n}}{\mu+{n}}\right) \\ $$ $$=\mathrm{ln}\left(\frac{\epsilon+\frac{{n}}{\mu}}{\mathrm{1}+\frac{{n}}{\mu}}\right) \\ $$ $$ \\ $$ $${k}=\frac{{n}}{\mu} \\ $$ $${A}=\mathrm{ln}\left(\frac{\epsilon+{k}}{\mathrm{1}+{k}}\right) \\ $$ $$ \\ $$ $$\mathrm{Similar}\:\mathrm{to}\:\mathrm{original}\:\mathrm{post}: \\ $$ $$\epsilon>\mathrm{1},\:{A}>\mathrm{0} \\ $$ $$\epsilon<\mathrm{1},\:{A}<\mathrm{0} \\ $$ $$\epsilon=\mathrm{1},\:{A}=\mathrm{0} \\ $$ $$ \\ $$

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