# ln-x-4-dx-

Question Number 68699 by Rio Michael last updated on 15/Sep/19
$$\int\:{ln}\left({x}\:+\:\mathrm{4}\right)\:{dx}\:= \\$$
Commented by mathmax by abdo last updated on 15/Sep/19
$${let}\:{I}\:=\int{ln}\left({x}+\mathrm{4}\right){dx}\:\:{changement}\:{x}+\mathrm{4}={t}\:{give} \\$$$${I}\:=\int{lnt}\:{dt}\:={tln}\left({t}\right)−{t}\:+{c}\:=\left({x}+\mathrm{4}\right){ln}\left({x}+\mathrm{4}\right)−{x}−\mathrm{4}\:+{c} \\$$
Commented by Cmr 237 last updated on 15/Sep/19
$$\mathrm{posons}\:\mathrm{ln}\left(\mathrm{x}+\mathrm{4}\right)=\mathrm{t}\:\mathrm{on} \\$$$$\mathrm{x}=\mathrm{e}^{\mathrm{t}} −\mathrm{4},\mathrm{dx}=\mathrm{e}^{\mathrm{t}} \mathrm{dt}\:\mathrm{ainsi}, \\$$$$\int\mathrm{ln}\left(\mathrm{x}+\mathrm{4}\right)\mathrm{dx}=\int\mathrm{te}^{\mathrm{t}} \mathrm{dt} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\mathrm{te}^{\mathrm{t}} \right]−\int\mathrm{e}^{\mathrm{t}} \mathrm{dt} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{t}−\mathrm{1}\right)\mathrm{e}^{\mathrm{t}} +\mathrm{c}\:,\mathrm{c}\in\mathbb{R} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{x}+\mathrm{4}\right)\left[\mathrm{ln}\left(\mathrm{x}+\mathrm{4}\right)−\mathrm{1}\right]+\mathrm{c} \\$$
Answered by Prithwish sen last updated on 15/Sep/19
$$\mathrm{Applying}\:\mathrm{by}\:\mathrm{part} \\$$$$=\mathrm{x}.\mathrm{ln}\left(\mathrm{x}+\mathrm{4}\right)−\int\frac{\mathrm{x}\:\mathrm{dx}}{\mathrm{x}+\mathrm{4}}\:\: \\$$$$=\:\mathrm{xln}\left(\mathrm{x}+\mathrm{4}\right)−\left[\int\mathrm{dx}\:−\mathrm{4}\int\frac{\mathrm{dx}}{\mathrm{x}+\mathrm{4}}\right] \\$$$$=\:\boldsymbol{\mathrm{xl}}\left(\boldsymbol{\mathrm{x}}+\mathrm{4}\right)−\boldsymbol{\mathrm{x}}\:+\:\mathrm{4}\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\mathrm{4}\right)\:+\boldsymbol{\mathrm{c}}\: \\$$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\$$
Commented by Rio Michael last updated on 15/Sep/19
$${i}\:{think}\:{perfect},{thanks}\:{sir} \\$$