Question Number 48495 by maxmathsup by imad last updated on 24/Nov/18
$$\left.\mathrm{1}\right){calculate}\:\:{I}\:=\int\:\frac{{ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}}{dt} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}}{dt} \\ $$
Commented by maxmathsup by imad last updated on 25/Nov/18
![by parts u^′ =(1/(1+t)) and v=ln(1+t) ⇒ I =ln^2 (1+t)−∫ ((ln(t))/(1+t))dt =ln^2 (1+t)−I ⇒ 2I =ln^2 (1+t) ⇒ I =(1/2)ln^2 (1+t)+c 2) ∫_0 ^1 ((ln(1+t))/(1+t))dt =[(1/2)ln^2 (1+t)]_0 ^1 =(1/2)ln^2 (2).](https://www.tinkutara.com/question/Q48555.png)
$${by}\:{parts}\:{u}^{'} =\frac{\mathrm{1}}{\mathrm{1}+{t}}\:{and}\:{v}={ln}\left(\mathrm{1}+{t}\right)\:\Rightarrow\:{I}\:={ln}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)−\int\:\frac{{ln}\left({t}\right)}{\mathrm{1}+{t}}{dt}\:={ln}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)−{I}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:={ln}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)\:\Rightarrow\:{I}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)+{c} \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}}{dt}\:=\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right). \\ $$$$ \\ $$
Answered by Abdulhafeez Abu qatada last updated on 25/Nov/18
