What is the integral of (1-e^{-10t}-10te^{-10t}) ?

The integral of (1-e^{-10t}-10te^{-10t}) is t+e^{-10t}t+1/5 e^{-10t}+C

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integral of (1-e^{-10t}-10te^{-10t})

\int (1 - e^{- 10 t} - 10 t e^{- 10 t}) d t

t + e^{- 10 t} t + \frac{1}{5} e^{- 10 t} + C

Solution steps

\int 1 - e^{- 10 t} - 10 t e^{- 10 t} 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 1 d t - \int e^{- 10 t} d t - \int 10 t e^{- 10 t} d t

\int 1 d t = t

\int e^{- 10 t} d t = - \frac{1}{10} e^{- 10 t}

\int 10 t e^{- 10 t} d t = - e^{- 10 t} t - \frac{1}{10} e^{- 10 t}

= t - (- \frac{1}{10} e^{- 10 t}) - (- e^{- 10 t} t - \frac{1}{10} e^{- 10 t})

Simplify

t - (- \frac{1}{10} e^{- 10 t}) - (- e^{- 10 t} t - \frac{1}{10} e^{- 10 t}) : t + e^{- 10 t} t + \frac{1}{5} e^{- 10 t}

= t + e^{- 10 t} t + \frac{1}{5} e^{- 10 t}

Add a constant to the solution

= t + e^{- 10 t} t + \frac{1}{5} e^{- 10 t} + C

y = 3 x, y = 3, x = 0

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

\frac{d}{d x} (arccsc (e^{x}))

\frac{\partial}{\partial y} (y^{4} cos (2 x))

y = x^{0.85}

What is the integral of (1-e^{-10t}-10te^{-10t}) ?
The integral of (1-e^{-10t}-10te^{-10t}) is t+e^{-10t}t+1/5 e^{-10t}+C