What is the integral of 700te^u(-8/(2t)) ?

The integral of 700te^u(-8/(2t)) is -2800e^u+C

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integral of 700te^u(-8/(2t))

\int 700 t e^{u} (- \frac{8}{2 t}) d u

- 2800 e^{u} + C

Solution steps

\int 700 t e^{u} (- \frac{8}{2 t}) d u

Take the constant out:

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

= (- \frac{8}{2 t}) \cdot 700 t \cdot \int e^{u} d u

(- \frac{8}{2 t}) \cdot 700 t = - 2800

= (- 2800) \cdot \int e^{u} d u

Use the common integral:

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

= (- 2800) e^{u}

Simplify

= - 2800 e^{u}

Add a constant to the solution

= - 2800 e^{u} + C

\frac{d}{d x} (tan^{c o s (x)} (x))

\frac{\partial}{\partial x} (x y sin (z))

\frac{d}{d x} (4 tan (x) + cot (x))

\frac{\partial}{\partial x} (5 x^{3} y)

(- 41.38, 4.9), (2.38, 3.7)

What is the integral of 700te^u(-8/(2t)) ?
The integral of 700te^u(-8/(2t)) is -2800e^u+C