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
![posons ln(x+4)=t on x=e^t −4,dx=e^t dt ainsi, ∫ln(x+4)dx=∫te^t dt =[te^t ]−∫e^t dt =(t−1)e^t +c ,c∈R =(x+4)[ln(x+4)−1]+c](Q68829.png)
$$\mathrm{posons}\:\mathrm{ln}\left(\mathrm{x}+\mathrm{4}\right)=\mathrm{t}\:\mathrm{on} \\ $$$$\mathrm{x}=\mathrm{e}^{\mathrm{t}} −\mathrm{4},\mathrm{dx}=\mathrm{e}^{\mathrm{t}} \mathrm{dt}\:\mathrm{ainsi}, \\ $$$$\int\mathrm{ln}\left(\mathrm{x}+\mathrm{4}\right)\mathrm{dx}=\int\mathrm{te}^{\mathrm{t}} \mathrm{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\mathrm{te}^{\mathrm{t}} \right]−\int\mathrm{e}^{\mathrm{t}} \mathrm{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{t}−\mathrm{1}\right)\mathrm{e}^{\mathrm{t}} +\mathrm{c}\:,\mathrm{c}\in\mathbb{R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{x}+\mathrm{4}\right)\left[\mathrm{ln}\left(\mathrm{x}+\mathrm{4}\right)−\mathrm{1}\right]+\mathrm{c} \\ $$
Answered by Prithwish sen last updated on 15/Sep/19
![Applying by part =x.ln(x+4)−∫((x dx)/(x+4)) = xln(x+4)−[∫dx −4∫(dx/(x+4))] = xl(x+4)−x + 4ln(x+4) +c please check.](Q68706.png)
$$\mathrm{Applying}\:\mathrm{by}\:\mathrm{part} \\ $$$$=\mathrm{x}.\mathrm{ln}\left(\mathrm{x}+\mathrm{4}\right)−\int\frac{\mathrm{x}\:\mathrm{dx}}{\mathrm{x}+\mathrm{4}}\:\: \\ $$$$=\:\mathrm{xln}\left(\mathrm{x}+\mathrm{4}\right)−\left[\int\mathrm{dx}\:−\mathrm{4}\int\frac{\mathrm{dx}}{\mathrm{x}+\mathrm{4}}\right] \\ $$$$=\:\boldsymbol{\mathrm{xl}}\left(\boldsymbol{\mathrm{x}}+\mathrm{4}\right)−\boldsymbol{\mathrm{x}}\:+\:\mathrm{4}\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\mathrm{4}\right)\:+\boldsymbol{\mathrm{c}}\: \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$
Commented by Rio Michael last updated on 15/Sep/19
$${i}\:{think}\:{perfect},{thanks}\:{sir} \\ $$