# A-dx-x-n-A-ln-x-n-ln-n-ln-n-ln-n-n-ln-n-n-n-A-dx-x-n-ln-n-1-For-a-b-c-i-if-a-gt-b-c-gt-1-ii

Question Number 3017 by Filup last updated on 03/Dec/15
$${A}=\int_{\mu} ^{\:\mu+\epsilon} \:\frac{{dx}}{{x}+{n}} \\$$$$\\$$$${A}=\mathrm{ln}\left({x}+{n}\right)\:\mid_{\mu} ^{\mu+\epsilon} \\$$$$=\mathrm{ln}\left(\mu+\epsilon+{n}\right)−\mathrm{ln}\left(\mu+{n}\right) \\$$$$=\mathrm{ln}\left(\frac{\mu+\epsilon+{n}}{\mu+{n}}\right) \\$$$$=\mathrm{ln}\left(\frac{\mu+{n}}{\mu+{n}}+\frac{\epsilon}{\mu+{n}}\right) \\$$$$\\$$$$\therefore{A}=\int_{\mu} ^{\:\mu+\epsilon} \:\frac{{dx}}{{x}+{n}}=\mathrm{ln}\left(\frac{\epsilon}{\mu+{n}}+\mathrm{1}\right) \\$$$$\\$$$$\mathrm{For}\:\:\frac{{a}}{{b}}={c} \\$$$$\left.{i}\right)\:\mathrm{if}\:\:{a}>{b},\:{c}>\mathrm{1} \\$$$$\left.{ii}\right)\:\mathrm{if}\:\:\:{a}<{b},\:{c}<\mathrm{1} \\$$$$\left.{iii}\right)\:\mathrm{if}\:\:{a}={b},\:{c}=\mathrm{1} \\$$$$\\$$$$\mathrm{Case}\:\left({i}\right) \\$$$$\epsilon>\mu+{n} \\$$$$\frac{\epsilon}{\mu+{n}}>\mathrm{1} \\$$$$\therefore{A}>\mathrm{ln}\left(\mathrm{2}\right) \\$$$$\\$$$$\mathrm{Case}\:\left({ii}\right) \\$$$$\epsilon<\mu+{n} \\$$$$\frac{\epsilon}{\mu+{n}}<\mathrm{1} \\$$$$\therefore{A}<\mathrm{ln}\left(\mathrm{2}\right) \\$$$$\\$$$$\mathrm{Case}\:\left({iii}\right) \\$$$$\epsilon=\mu+{n} \\$$$$\frac{\epsilon}{\mu+{n}}=\mathrm{1} \\$$$${A}=\mathrm{ln}\left(\mathrm{2}\right) \\$$$$\\$$$$\mathrm{What}\:\mathrm{other}\:\mathrm{observations}\:\mathrm{can}\:\mathrm{you}\:\mathrm{find} \\$$$$\mathrm{regarding}\:\mathrm{similar}\:\mathrm{logarithmic}\:\mathrm{functions}? \\$$
Commented by Filup last updated on 03/Dec/15
$$\mathrm{Maybe}\:\mathrm{look}\:\mathrm{at}\:\mathrm{cases}\:\mathrm{of}\:\int_{\mu} ^{\:\mu\epsilon} ,\:\int_{\mu} ^{\:\mu^{\epsilon} } ,\:{etc}. \\$$
Commented by 123456 last updated on 03/Dec/15
$$\mu>−{n}\vee\mu+\epsilon<−{n} \\$$
Commented by Filup last updated on 03/Dec/15
$$\mathrm{Please}\:\mathrm{elaborate}? \\$$
Answered by Filup last updated on 03/Dec/15
$$\int_{\mu} ^{\mu\epsilon} \:\frac{{dx}}{{x}+{n}}=\mathrm{ln}\left(\frac{\mu\epsilon+{n}}{\mu+{n}}\right) \\$$$$=\mathrm{ln}\left(\frac{\epsilon+\frac{{n}}{\mu}}{\mathrm{1}+\frac{{n}}{\mu}}\right) \\$$$$\\$$$${k}=\frac{{n}}{\mu} \\$$$${A}=\mathrm{ln}\left(\frac{\epsilon+{k}}{\mathrm{1}+{k}}\right) \\$$$$\\$$$$\mathrm{Similar}\:\mathrm{to}\:\mathrm{original}\:\mathrm{post}: \\$$$$\epsilon>\mathrm{1},\:{A}>\mathrm{0} \\$$$$\epsilon<\mathrm{1},\:{A}<\mathrm{0} \\$$$$\epsilon=\mathrm{1},\:{A}=\mathrm{0} \\$$$$\\$$