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

x \to - 1 lim (\frac{2 x + 2}{x + 1})

f (x) = \frac{2}{x - 6}, (8, 1)

\int x^{n} I n x d x

\int \frac{- 10 x ^{2}}{( x + 1 ) ^{3}} d x

25 x^{8} - 34 x^{5} + 2^{3} - x

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