What is the integral of (t+t^2)e^{-t} ?

The integral of (t+t^2)e^{-t} is -3e^{-t}t-3e^{-t}-e^{-t}t^2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (t+t^2)e^{-t}

\int (t + t^{2}) e^{- t} d t

- 3 e^{- t} t - 3 e^{- t} - e^{- t} t^{2} + C

Solution steps

\int (t + t^{2}) e^{- t} d t

Apply u-substitution

: \int - e^{u} (- u + u^{2}) d u

= \int - e^{u} (- u + u^{2}) d u

Take the constant out:

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

= - \int e^{u} (- u + u^{2}) d u

Expand

e^{u} (- u + u^{2}) : - e^{u} u + e^{u} u^{2}

= - \int - e^{u} u + e^{u} u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int e^{u} u d u + \int e^{u} u^{2} d u)

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

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

= - (- (e^{u} u - e^{u}) + u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - t

= - (- (e^{- t} (- t) - e^{- t}) + (- t)^{2} e^{- t} - 2 (e^{- t} (- t) - e^{- t}))

Simplify

- (- (e^{- t} (- t) - e^{- t}) + (- t)^{2} e^{- t} - 2 (e^{- t} (- t) - e^{- t})) : - 3 e^{- t} t - 3 e^{- t} - e^{- t} t^{2}

= - 3 e^{- t} t - 3 e^{- t} - e^{- t} t^{2}

Add a constant to the solution

= - 3 e^{- t} t - 3 e^{- t} - e^{- t} t^{2} + C

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

\int \frac{3 x - 1}{x ^{2} - 5 x} d x

\int \frac{( x + 1 )}{x ^{2} + 2 x + 4} d x

\int e^{x} \frac{1}{1 + e ^{2 x}} d x

\int 3 x + 6 cos (x) d x

What is the integral of (t+t^2)e^{-t} ?
The integral of (t+t^2)e^{-t} is -3e^{-t}t-3e^{-t}-e^{-t}t^2+C