Question Number 33327 by prof Abdo imad last updated on 14/Apr/18
$${find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\mathrm{3}\:+{e}^{−{x}} } \\ $$
Commented by math khazana by abdo last updated on 19/Apr/18
![let put I = ∫_0 ^1 (dx/(3 +e^(−x) )) changement e^x =t give I = ∫_1 ^e (1/(3 +(1/t))) (dt/t) = ∫_1 ^e (dt/(3t +1)) =[(1/3)ln(3t+1)]_1 ^e = (1/3)( ln(3e+1) −ln(4)) =(1/3)(ln(3e+1) −2ln(2)) .](https://www.tinkutara.com/question/Q33567.png)
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{\mathrm{3}\:+{e}^{−{x}} }\:\:\:\:\:{changement}\:\:{e}^{{x}} \:={t}\:{give} \\ $$$${I}\:=\:\int_{\mathrm{1}} ^{{e}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{3}\:+\frac{\mathrm{1}}{{t}}}\:\frac{{dt}}{{t}}\:\:=\:\int_{\mathrm{1}} ^{{e}} \:\:\:\:\:\frac{{dt}}{\mathrm{3}{t}\:+\mathrm{1}} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{3}}{ln}\left(\mathrm{3}{t}+\mathrm{1}\right)\right]_{\mathrm{1}} ^{{e}} \:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\:{ln}\left(\mathrm{3}{e}+\mathrm{1}\right)\:−{ln}\left(\mathrm{4}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left({ln}\left(\mathrm{3}{e}+\mathrm{1}\right)\:−\mathrm{2}{ln}\left(\mathrm{2}\right)\right)\:. \\ $$