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

The integral of te^{3t} is 1/9 (e^{3t}*3t-e^{3t})+C

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integral of te^{3t}

\int t e^{3 t} d t

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

Solution steps

\int t e^{3 t} d t

Apply u-substitution

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

= \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = 3 t

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

Add a constant to the solution

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 t e^{3 t}

f (x) = ln (9 sin^{2} (x))

\int \frac{3 x - 1}{x ^{2} - 4} d x

- 3 (sin (4 (t - \frac{5 π}{4})))

What is the integral of te^{3t} ?
The integral of te^{3t} is 1/9 (e^{3t}*3t-e^{3t})+C