Question Number 40129 by maxmathsup by imad last updated on 16/Jul/18
$${calculate}\:{I}\:=\:\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}^{\mathrm{2}} }{dt} \\ $$
Answered by maxmathsup by imad last updated on 21/Jul/18
![by parts I = [−(1/t)ln(1+t)]_1 ^2 +∫_1 ^2 (dt/(t(1+t))) =ln(2)−(1/2)ln(3) + ∫_1 ^2 ((1/t) −(1/(t+1)))dt =ln(2)−(1/2)ln(3) +[ln∣(t/(t+1))∣]_1 ^2 =ln(2)−(1/2)ln(3)+ln((2/3))−ln((1/2)) I=3ln(2)−(3/2)ln(3)](https://www.tinkutara.com/question/Q40428.png)
$${by}\:{parts}\:\:{I}\:=\:\left[−\frac{\mathrm{1}}{{t}}{ln}\left(\mathrm{1}+{t}\right)\right]_{\mathrm{1}} ^{\mathrm{2}} \:+\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{dt}}{{t}\left(\mathrm{1}+{t}\right)} \\ $$$$={ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{3}\right)\:+\:\int_{\mathrm{1}} ^{\mathrm{2}} \left(\frac{\mathrm{1}}{{t}}\:−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt} \\ $$$$={ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{3}\right)\:+\left[{ln}\mid\frac{{t}}{{t}+\mathrm{1}}\mid\right]_{\mathrm{1}} ^{\mathrm{2}} \:={ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{3}\right)+{ln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${I}=\mathrm{3}{ln}\left(\mathrm{2}\right)−\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\mathrm{3}\right) \\ $$