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

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

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integral of (te^{-t})

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

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

Solution steps

\int t e^{- t} d t

Apply Integration By Parts

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

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

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

= - e^{- t} t - e^{- t}

Add a constant to the solution

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

\frac{d y}{d x} = sin (2 x)

y^{2} = x + 1, y + x = 1

\frac{\partial}{\partial x} (e^{5 x})

sin (2 x), \frac{π}{2}

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