What is the integral of te^{8t} ?

The integral of te^{8t} is 1/64 (e^{8t}*8t-e^{8t})+C

integral of te^{8t}

\int t e^{8 t} d t

\frac{1}{64} (e^{8 t} \cdot 8 t - e^{8 t}) + C

Solution steps

\int t e^{8 t} d t

Apply u-substitution

= \int \frac{e ^{u} u}{64} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int e^{u} u d u

Apply Integration By Parts

= \frac{1}{64} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{64} (e^{u} u - e^{u})

Substitute back

u = 8 t

= \frac{1}{64} (e^{8 t} \cdot 8 t - e^{8 t})

Add a constant to the solution

= \frac{1}{64} (e^{8 t} \cdot 8 t - e^{8 t}) + C

f (x) = 5 x^{- \frac{3}{5}}

y = x^{3} - 7 x^{2} + 8 x + 1

2 y^{^{''}} + y^{^{'}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

\frac{\partial}{\partial y} (x \cdot e^{x \cdot y})

\frac{\partial}{\partial x} (- \frac{x}{( x ^{2} + y ^{2} )})

What is the integral of te^{8t} ?
The integral of te^{8t} is 1/64 (e^{8t}*8t-e^{8t})+C