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

The integral of-4te^{-4t} is e^{-4t}t+1/4 e^{-4t}+C

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integral of-4te^{-4t}

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

e^{- 4 t} t + \frac{1}{4} e^{- 4 t} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{e ^{u} u}{16} d u

= - 4 \cdot \int \frac{e ^{u} u}{16} d u

Take the constant out:

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

= - 4 \cdot \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= - 4 \cdot \frac{1}{16} (e^{u} u - \int e^{u} d u)

\int e^{u} d u = e^{u}

= - 4 \cdot \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = - 4 t

= - 4 \cdot \frac{1}{16} (e^{- 4 t} (- 4 t) - e^{- 4 t})

Simplify

- 4 \cdot \frac{1}{16} (e^{- 4 t} (- 4 t) - e^{- 4 t}) : e^{- 4 t} t + \frac{1}{4} e^{- 4 t}

= e^{- 4 t} t + \frac{1}{4} e^{- 4 t}

Add a constant to the solution

= e^{- 4 t} t + \frac{1}{4} e^{- 4 t} + C

\int 5 tan (x) d x

4 y^{^{''}} + 3 y = 0

\int 8 x - 3 d x

\int_{1}^{2} \frac{x ^{2} - x - 5}{x + 2} d x

y = \frac{4}{x}, y = \frac{4}{x ^{2}}, x = 3

What is the integral of-4te^{-4t} ?
The integral of-4te^{-4t} is e^{-4t}t+1/4 e^{-4t}+C