Question Number 87769 by john santu last updated on 06/Apr/20
$$\int\:\frac{\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} +\mathrm{1}\right)}{\mathrm{e}^{−\mathrm{x}} +\mathrm{1}}\:\mathrm{dx}\: \\ $$
Answered by TANMAY PANACEA. last updated on 06/Apr/20
$$\int\frac{{e}^{{x}} {ln}\left({e}^{{x}} +\mathrm{1}\right)}{{e}^{{x}} +\mathrm{1}}{dx} \\ $$$${t}={ln}\left({e}^{{x}} +\mathrm{1}\right)\rightarrow\frac{{dt}}{{dx}}=\frac{{e}^{{x}} }{{e}^{{x}} +\mathrm{1}} \\ $$$$\int{tdt}=\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$$$\frac{\left\{{ln}\left({e}^{{x}} +\mathrm{1}\right)\right\}^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$
Answered by john santu last updated on 06/Apr/20
