What is the integral of t*e^{-at} ?

The integral of t*e^{-at} is 1/(a^2)(-ae^{-at}t-e^{-at})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of t*e^{-at}

\int t \cdot e^{- a t} d t

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - a t

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

Simplify

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

Add a constant to the solution

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

\int \frac{x ^{2} + 2 x + 4}{x} d x

\int 10 x^{- 2} d x

\int y^{2} d y

\int \frac{- 18 x ^{2} + 12}{x ^{3} - 2 x - 8} d x

\int cot^{2} (6 x^{2}) x d x

What is the integral of t*e^{-at} ?
The integral of t*e^{-at} is 1/(a^2)(-ae^{-at}t-e^{-at})+C