What is the integral of 4e^{6x} ?

The integral of 4e^{6x} is 2/3 e^{6x}+C

integral of 4e^{6x}

\int 4 e^{6 x} d x

\frac{2}{3} e^{6 x} + C

Solution steps

\int 4 e^{6 x} d x

Take the constant out:

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

= 4 \cdot \int e^{6 x} d x

Apply u-substitution

= 4 \cdot \int e^{u} \frac{1}{6} d u

Take the constant out:

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

= 4 \cdot \frac{1}{6} \cdot \int e^{u} d u

Use the common integral:

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

= 4 \cdot \frac{1}{6} e^{u}

Substitute back

u = 6 x

= 4 \cdot \frac{1}{6} e^{6 x}

Simplify

4 \cdot \frac{1}{6} e^{6 x} : \frac{2}{3} e^{6 x}

= \frac{2}{3} e^{6 x}

Add a constant to the solution

= \frac{2}{3} e^{6 x} + C

\frac{\partial}{\partial x} (cos^{6} (x^{7} y^{8}))

\int_{\frac{1}{2}}^{1} (x^{- 3} - 6) d x

3 x d x + 4 y d y = 0

\frac{d}{d x} (- 3 x - 2 3 x^{2})

s l o p e (- 5, 5), (5, - 2)