What is the integral of (1-x)/(e^x) ?

The integral of (1-x)/(e^x) is -(1-x)/(e^x)+1/(e^x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1-x)/(e^x)

\int \frac{1 - x}{e ^{x}} d x

- \frac{1 - x}{e ^{x}} + \frac{1}{e ^{x}} + C

Solution steps

\int \frac{1 - x}{e ^{x}} d x

Apply u-substitution

: \int \frac{1 - ln ( u )}{u ^{2}} d u

= \int \frac{1 - ln ( u )}{u ^{2}} d u

Apply Integration By Parts

: - \frac{1 - ln ( u )}{u} - \int \frac{1}{u ^{2}} d u

= - \frac{1 - ln ( u )}{u} - \int \frac{1}{u ^{2}} d u

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

= - \frac{1 - ln ( u )}{u} - (- \frac{1}{u})

Substitute back

u = e^{x}

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

Simplify

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

= - \frac{1 - x}{e ^{x}} + \frac{1}{e ^{x}}

Add a constant to the solution

= - \frac{1 - x}{e ^{x}} + \frac{1}{e ^{x}} + C

\int t sec^{2} (2 t) d t

\frac{\partial}{\partial y} (\frac{y}{( x + y ) ^{2}})

\frac{d}{d x} ((3 x^{2} + ln (x)) (5 x^{3})^{x})

f (t) = t^{\frac{5}{3}} e^{t}

\int_{0}^{\infty} e^{- x} cos (x) d x

What is the integral of (1-x)/(e^x) ?
The integral of (1-x)/(e^x) is -(1-x)/(e^x)+1/(e^x)+C