부분적분 정적분

$x = f' ( x ) , 1 - \ln x = g ( x )$라 하면 $\displaystyle \int _{1} ^{e} f' ( x ) g ( x ) dx = \left[ f ( x ) g ( x ) \right] _{1} ^{e} - \int _{1} ^{e} f ( x ) g' ( x ) dx$ 임을 이용하자. $\displaystyle \int _{1} ^{e} x ( 1 - \ln x ) dx = \left[ \frac{1}{2} x ^{2} ( 1 - \ln x ) \right] _{1} ^{e} - \int _{1} ^{e} \left( \frac{1}{2} x ^{2} \right) \left( - \frac{1}{x} \right) dx$ $\displaystyle = \left( \frac{1}{2} e ^{2} - \frac{1}{2} \right) + \int _{1} ^{e} \frac{1}{2} x dx$ $\displaystyle = \left( \frac{1}{2} e ^{2} - \frac{1}{2} \right) + \left[ \frac{1}{4} x ^{2} \right] _{1} ^{e}$ $\displaystyle = \left( \frac{1}{2} e ^{2} - \frac{1}{2} \right) + \left( \frac{1}{4} e ^{2} - \frac{1}{4} \right)$ $\displaystyle = \frac{1}{4} e ^{2} - \frac{3}{4}$ $\displaystyle = \frac{1}{4} ( e ^{2} - 3 )$