로그 함수 정적분

두 점 ${\mathrm{P}}$, ${\mathrm{Q}}$의 $x$좌표를 구하면 $e ^{x} = t$에서 $x = \ln t$ $e ^{x} = t + 2$에서 $x = \ln \left( t + 2 \right)$ $f \left( t \right) = \ln \left( t + 2 \right) - \ln t$이므로 $\displaystyle \int _{\frac{2}{3}} ^{2} \frac{\ln \left( t + 2 \right) - \ln t}{t ^{2}} dt = \int _{\frac{2}{3}} ^{2} \frac{1}{t ^{2}} \ln \left( 1 + \frac{2}{t} \right) dt$ $\displaystyle 1 + \frac{2}{t} = u$라 하면 $\displaystyle - \frac{2}{t ^{2}} = \frac{du}{dt}$ $\displaystyle t = \frac{2}{3}$일 때 $u = 4$ $t = 2$일 때 $u = 2$ $\displaystyle \int _{\frac{2}{3}} ^{2} \frac{1}{t ^{2}} \ln \left( 1 + \frac{2}{t} \right) dt$ $\displaystyle = \int _{4} ^{2} {} \left( - \frac{1}{2} \ln u \right) du$ $\displaystyle = \frac{1}{2} \int _{2} ^{4} {} \ln u du$ $\displaystyle = \frac{1}{2} \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} u \ln u - u \end{array} \right] _{2} ^{4}$ $\displaystyle = \frac{1}{2} \left\{ \left( 4 \ln 4 - 4 \right) - \left( 2 \ln 2 - 2 \right) \right\}$ $\displaystyle = \frac{1}{2} \left\{ \left( 8 \ln 2 - 4 \right) - \left( 2 \ln 2 - 2 \right) \right\}$ $\displaystyle = \frac{1}{2} \left( 6 \ln 2 - 2 \right) = 3 \ln 2 - 1$ 따라서 $\displaystyle \int _{\frac{2}{3}} ^{2} \frac{f \left( t \right)}{t ^{2}} dt = - 1 + 3 \ln 2$