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

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

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integral of e^{2t}t

\int e^{2 t} t d t

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

Solution steps

\int e^{2 t} 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

f (x) = 3 x^{2} - 5 x - 1 + e^{- x}, at x = 0

\int 2 x^{2} - 10 x d x

\int \frac{40}{16 - x ^{2}} d x

\frac{d ^{2}}{d x ^{2}} (\frac{8}{9 x ^{5}})

\frac{\partial}{\partial z} (\frac{x - y}{y + z})

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