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} \cdot 2 t - e^{2 t})

Simplify

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

x \to 0 - lim (\frac{3}{e ^{x} + 1})

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

\frac{\partial}{\partial z} (e^{x} sin (y))

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 8} e^{5 t}

\int 7 x e^{5 x} d x

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