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

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

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integral of te^{2t}

\int t e^{2 t} d t

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

Solution steps

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

Add a constant to the solution

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

\int sin^{4} (3 t) d t

\frac{\partial}{\partial x} (- \frac{2}{1 - 2 x})

\frac{\partial}{\partial x} (x^{2} (y - x))

9 x y

\frac{d}{d x} (\frac{x + y + 1}{x - 1})

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