What is the integral of 18te^{3t} ?

The integral of 18te^{3t} is 2(3e^{3t}t-e^{3t})+C

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integral of 18te^{3t}

\int 18 t e^{3 t} d t

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

Solution steps

\int 18 t e^{3 t} d t

Take the constant out:

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

= 18 \cdot \int t e^{3 t} d t

Apply u-substitution

= 18 \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

= 18 \cdot \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

= 18 \cdot \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

= 18 \cdot \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = 3 t

= 18 \cdot \frac{1}{9} (e^{3 t} \cdot 3 t - e^{3 t})

Simplify

18 \cdot \frac{1}{9} (e^{3 t} \cdot 3 t - e^{3 t}) : 2 (3 e^{3 t} t - e^{3 t})

= 2 (3 e^{3 t} t - e^{3 t})

Add a constant to the solution

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

What is the integral of 18te^{3t} ?
The integral of 18te^{3t} is 2(3e^{3t}t-e^{3t})+C