What is expand (2019+ln(x-1)(x-1))/(e^x) ?

The solution to expand (2019+ln(x-1)(x-1))/(e^x) is (2019)/(e^x)+(xln(x-1))/(e^x)-(ln(x-1))/(e^x)

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expand (2019+ln(x-1)(x-1))/(e^x)

expand

\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}}

Solution steps

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

Expand

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}}

Apply the fraction rule:

\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}}

(x + \frac{2}{3})^{3}

\frac{x ^{(2)} + x + 2}{x ^{(2)} ( x ^{(2)} + 3 ) ^{(2)}}

\frac{2 x - x ^{2}}{3 ( 2 - 2 x ) ^{2}}

(3 n - 3) 2

\frac{12 - 25}{2} - \frac{2}{3} \cdot (2 \cdot (- 2.9342) + 0.73683)

What is expand (2019+ln(x-1)(x-1))/(e^x) ?
The solution to expand (2019+ln(x-1)(x-1))/(e^x) is (2019)/(e^x)+(xln(x-1))/(e^x)-(ln(x-1))/(e^x)