What is the integral of 4e^{-t}t ?

The integral of 4e^{-t}t is 4(-e^{-t}t-e^{-t})+C

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integral of 4e^{-t}t

\int 4 e^{- t} t d t

4 (- e^{- t} t - e^{- t}) + C

Solution steps

\int 4 e^{- t} t d t

Take the constant out:

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

= 4 \cdot \int e^{- t} t d t

Apply Integration By Parts

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

= 4 (- e^{- t} t - \int - e^{- t} d t)

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

= 4 (- e^{- t} t - e^{- t})

Add a constant to the solution

= 4 (- e^{- t} t - e^{- t}) + C

\int sec (8 x) d x

f (x) = 6 sec (x)

\frac{\partial}{\partial x} (x^{3} + 8 x^{2} y + 12 y^{2})

\frac{d y}{d x} + 8 y = e^{7 x}

\frac{d}{d x} (e^{x} (11 x^{2} - 22 x + 22))

What is the integral of 4e^{-t}t ?
The integral of 4e^{-t}t is 4(-e^{-t}t-e^{-t})+C