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

The integral of (3te^{3t}) is e^{3t}t-1/3 e^{3t}+C

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integral of (3te^{3t})

\int (3 t e^{3 t}) d t

e^{3 t} t - \frac{1}{3} e^{3 t} + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u} u}{9} d u

= 3 \cdot \int \frac{e ^{u} u}{9} d u

Take the constant out:

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

= 3 \cdot \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

= 3 \cdot \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = 3 t

= 3 \cdot \frac{1}{9} (e^{3 t} \cdot 3 t - e^{3 t})

Simplify

3 \cdot \frac{1}{9} (e^{3 t} \cdot 3 t - e^{3 t}) : e^{3 t} t - \frac{1}{3} e^{3 t}

= e^{3 t} t - \frac{1}{3} e^{3 t}

Add a constant to the solution

= e^{3 t} t - \frac{1}{3} e^{3 t} + C

\int_{- 2}^{1} x^{2} - 1 d x

\frac{\partial}{\partial x} (arcsin (2 x - 5))

x = y^{2}, x = 36

x \to 0 lim (cos^{2} (x) - 1)

\int \frac{x ^{2} + 3 x + 2}{x ( x ^{2} + 2 x + 2 )} d x

What is the integral of (3te^{3t}) ?
The integral of (3te^{3t}) is e^{3t}t-1/3 e^{3t}+C