What is the integral of 3te^{2t} ?

The integral of 3te^{2t} is 3/4 (2e^{2t}t-e^{2t})+C

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integral of 3te^{2t}

\int 3 t e^{2 t} d t

\frac{3}{4} (2 e^{2 t} t - e^{2 t}) + C

Solution steps

\int 3 t e^{2 t} d t

Take the constant out:

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

= 3 \cdot \int t e^{2 t} d t

Apply u-substitution

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

= 3 \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

= 3 \cdot \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

= 3 \cdot \frac{1}{4} (e^{u} u - e^{u})

Substitute back

u = 2 t

= 3 \cdot \frac{1}{4} (e^{2 t} \cdot 2 t - e^{2 t})

Simplify

3 \cdot \frac{1}{4} (e^{2 t} \cdot 2 t - e^{2 t}) : \frac{3}{4} (2 e^{2 t} t - e^{2 t})

= \frac{3}{4} (2 e^{2 t} t - e^{2 t})

Add a constant to the solution

= \frac{3}{4} (2 e^{2 t} t - e^{2 t}) + C

x \to 1 lim (- 6 + 4 x^{2})

2 x - 5 y = 0, y = 0, x = 5

\frac{d}{d t} (e^{- t})

\int 4 + 25 x^{2} d x

y = x^{3}, (1, 0)

What is the integral of 3te^{2t} ?
The integral of 3te^{2t} is 3/4 (2e^{2t}t-e^{2t})+C