What is the integral of-te^{-2t} ?

The integral of-te^{-2t} is -1/4 (-2e^{-2t}t-e^{-2t})+C

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integral of-te^{-2t}

\int - t e^{- 2 t} d t

- \frac{1}{4} (- 2 e^{- 2 t} t - e^{- 2 t}) + C

Solution steps

\int - t e^{- 2 t} d t

Take the constant out:

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

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

Apply u-substitution

: \int \frac{e ^{u} u}{4} d u

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - 2 t

= - \frac{1}{4} (e^{- 2 t} (- 2 t) - e^{- 2 t})

Simplify

= - \frac{1}{4} (- 2 e^{- 2 t} t - e^{- 2 t})

Add a constant to the solution

= - \frac{1}{4} (- 2 e^{- 2 t} t - e^{- 2 t}) + C

t^{2} \frac{d y}{d t} + y = 0, y (5) = - 1

\int 1. 8^{(0.045 x)} - 1 d x

\int - x^{2} e^{- x} d x

(1 - x^{2}) y^{^{''}} - 2 x y^{^{'}} = 1

x \to 0 lim (1 - 1 - x)

What is the integral of-te^{-2t} ?
The integral of-te^{-2t} is -1/4 (-2e^{-2t}t-e^{-2t})+C