What is the integral of-2e^{-t}+8t+6 ?

The integral of-2e^{-t}+8t+6 is 2e^{-t}+4t^2+6t+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-2e^{-t}+8t+6

\int - 2 e^{- t} + 8 t + 6 d t

2 e^{- t} + 4 t^{2} + 6 t + C

Solution steps

\int - 2 e^{- t} + 8 t + 6 d t

Apply the Sum Rule:

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

= - \int 2 e^{- t} d t + \int 8 t d t + \int 6 d t

\int 2 e^{- t} d t = - 2 e^{- t}

\int 8 t d t = 4 t^{2}

\int 6 d t = 6 t

= - (- 2 e^{- t}) + 4 t^{2} + 6 t

Simplify

= 2 e^{- t} + 4 t^{2} + 6 t

Add a constant to the solution

= 2 e^{- t} + 4 t^{2} + 6 t + C

\int x_{0} d x

\int \frac{1}{5 x ^{2} - 10 x + 6} d x

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

\int 3 x - 24 d x

\int (7 - 2 x) 9 + 16 x - 4 x^{2} d x

What is the integral of-2e^{-t}+8t+6 ?
The integral of-2e^{-t}+8t+6 is 2e^{-t}+4t^2+6t+C