What is the integral of (te^ti-4e^{-2t}j+8te^{t^2}k) ?

The integral of (te^ti-4e^{-2t}j+8te^{t^2}k) is (2je^{-2t}+4ke^{t^2})+i(-e^t+e^tt)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (te^ti-4e^{-2t}j+8te^{t^2}k)

\int (t e^{t} i - 4 e^{- 2 t} j + 8 t e^{t^{2}} k) d t

(2 j e^{- 2 t} + 4 k e^{t^{2}}) + i (- e^{t} + e^{t} t) + C

Solution steps

\int t e^{t} i - 4 e^{- 2 t} j + 8 t e^{t^{2}} k 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 t e^{t} i d t - \int 4 e^{- 2 t} j d t + \int 8 t e^{t^{2}} k d t

\int t e^{t} i d t = i (e^{t} t - e^{t})

\int 4 e^{- 2 t} j d t = - 2 j e^{- 2 t}

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

= i (e^{t} t - e^{t}) - (- 2 j e^{- 2 t}) + 4 k e^{t^{2}}

Simplify

i (e^{t} t - e^{t}) - (- 2 j e^{- 2 t}) + 4 k e^{t^{2}} : (2 j e^{- 2 t} + 4 k e^{t^{2}}) + i (- e^{t} + e^{t} t)

= (2 j e^{- 2 t} + 4 k e^{t^{2}}) + i (- e^{t} + e^{t} t)

Add a constant to the solution

= (2 j e^{- 2 t} + 4 k e^{t^{2}}) + i (- e^{t} + e^{t} t) + C

\frac{\partial}{\partial x} (\frac{2 x + 3}{4 y - 1})

(90, 62536), (950, 9000)

\frac{d}{d t} (\frac{A e ^{m t}}{B + C e ^{m t}})

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

\int (tan (\frac{x}{14}))^{5} d x

What is the integral of (te^ti-4e^{-2t}j+8te^{t^2}k) ?
The integral of (te^ti-4e^{-2t}j+8te^{t^2}k) is (2je^{-2t}+4ke^{t^2})+i(-e^t+e^tt)+C