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Question Number 40154 by maxmathsup by imad last updated on 16/Jul/18

$$findthevalueof{\int }_0^1\frac{ln\left(t\right)}{{(1+t)}^2}dt$$

Commented by maxmathsup by imad last updated on 16/Jul/18

$$letI={\int }_0^1\frac{ln\left(t\right)}{{(1+t)}^2}dtletintegratebyparts{u}^{{}^{\prime }}=\frac{1}{{(1+t)}^2}andv=ln\left(t\right)$$$$I={\left[(1-\frac{1}{1+t})ln\left(t\right)\right]}_0^1-{\int }_0^1(1-\frac{1}{1+t})\frac{dt}{t}$$$$=-{\int }_0^1\frac{dt}{1+t}=-{[ln\mid 1+t\mid ]}_0^1=-ln\left(2\right)$$$$I=-ln\left(2\right).$$

Answered by ajfour last updated on 16/Jul/18

$$I={\int }_0^1\frac{\ln t}{{(1+t)}^2}$$$$=-\frac{\ln t}{(1+t)}{\mid }_0^1+{\int }_0^1\frac{dt}{t(1+t)}$$$$=\underset{t\to 0}{\lim }[\left(\frac{\ln t}{1+t}\right)-\ln \left(\frac{t}{1+t}\right)]-\ln 2$$$$=\underset{t\to 0}{\lim }\left[\ln \frac{(1+t)t^{1+t}}{t}\right]-\ln 2$$$$=\underset{t\to 0}{\lim }\left[\ln (1+t)t^t\right]-\ln 2$$$$=\underset{t\to 0}{\lim ln}(1+t)+\underset{t\to 0}{\lim }\frac{\ln t}{\left(1/t\right)}-\ln 2$$$$=0+\underset{t\to 0}{\lim }\left(\frac{1/t}{-1/t^2}\right)-\ln 2$$$$I=-\ln 2.$$$$$$

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