What is the integral of (e^{-st}t) ?

The integral of (e^{-st}t) is 1/(s^2)(-se^{-st}t-e^{-st})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^{-st}t)

\int (e^{- s t} t) d t

\frac{1}{s ^{2}} (- s e^{- s t} t - e^{- s t}) + C

Solution steps

\int e^{- s t} t d t

Apply u-substitution

: \int \frac{e ^{u} u}{s ^{2}} d u

= \int \frac{e ^{u} u}{s ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{s ^{2}} \cdot \int e^{u} u d u

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

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

\int e^{u} d u = e^{u}

= \frac{1}{s ^{2}} (e^{u} u - e^{u})

Substitute back

u = - s t

= \frac{1}{s ^{2}} (e^{- s t} (- s t) - e^{- s t})

Simplify

= \frac{1}{s ^{2}} (- s e^{- s t} t - e^{- s t})

Add a constant to the solution

= \frac{1}{s ^{2}} (- s e^{- s t} t - e^{- s t}) + C

x^{2} + 1

\int \frac{6}{3 x} d x

\int \frac{1}{1 + v} d v

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

\int e^{- 2 z} d z

What is the integral of (e^{-st}t) ?
The integral of (e^{-st}t) is 1/(s^2)(-se^{-st}t-e^{-st})+C