What is the integral of te^{4t} ?

The integral of te^{4t} is 1/16 (e^{4t}*4t-e^{4t})+C

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integral of te^{4t}

\int t e^{4 t} d t

\frac{1}{16} (e^{4 t} \cdot 4 t - e^{4 t}) + C

Solution steps

\int t e^{4 t} d t

Apply u-substitution

: \int \frac{e ^{u} u}{16} d u

= \int \frac{e ^{u} u}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = 4 t

= \frac{1}{16} (e^{4 t} \cdot 4 t - e^{4 t})

Add a constant to the solution

= \frac{1}{16} (e^{4 t} \cdot 4 t - e^{4 t}) + C

\int \frac{8}{( 16 x ^{2} + 1 ) ^{2}} d x

\frac{\partial}{\partial y} (e^{x y z^{6}} y z^{6})

\frac{d}{d x} (7 x + 5 x)

1 - x

y^{^{''}} + 8 y^{^{'}} + 16 y = 1521 e^{2 t} cos (9 t)

What is the integral of te^{4t} ?
The integral of te^{4t} is 1/16 (e^{4t}*4t-e^{4t})+C