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

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

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integral of 2te^{-2t}

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

= 2 \cdot \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 2 \cdot \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = - 2 t

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

Simplify

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

= - e^{- 2 t} t - \frac{1}{2} e^{- 2 t}

Add a constant to the solution

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

\int (e^{2 x^{2}}) d x

\int \frac{sin ( ln ( x ^{3} ) )}{x} d x

a e^{3 x}

g (x) = (x + 2 x) e^{x}

\frac{1}{( s - 2 ) ^{2}}

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