展開する (2019+ln(x-1)(x-1))/(e^x)

解

展開する

\frac{2019 + ln ( x - 1 ) ( x - 1 )}{e ^{x}}

\frac{2019}{e ^{x}} + \frac{x ln ( x - 1 )}{e ^{x}} - \frac{ln ( x - 1 )}{e ^{x}}

解答ステップ

\frac{2019 + ln ( x - 1 ) ( x - 1 )}{e ^{x}}

拡張

2019 + ln (x - 1) (x - 1) : 2019 + x ln (x - 1) - ln (x - 1)

= \frac{2019 + x ln ( x - 1 ) - ln ( x - 1 )}{e ^{x}}

分数の規則を適用する:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{2019 + x ln ( x - 1 ) - ln ( x - 1 )}{e ^{x}} = \frac{2019}{e ^{x}} + \frac{x ln ( x - 1 )}{e ^{x}} - \frac{ln ( x - 1 )}{e ^{x}}

= \frac{2019}{e ^{x}} + \frac{x ln ( x - 1 )}{e ^{x}} - \frac{ln ( x - 1 )}{e ^{x}}