What is the integral of 2ye^{2z} ?

The integral of 2ye^{2z} is ye^{2z}+C

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integral of 2ye^{2z}

\int 2 y e^{2 z} d z

y e^{2 z} + C

Solution steps

\int 2 y e^{2 z} d z

Take the constant out:

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

= 2 y \cdot \int e^{2 z} d z

Apply u-substitution

: \int e^{u} \frac{1}{2} d u

= 2 y \cdot \int e^{u} \frac{1}{2} d u

Take the constant out:

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

= 2 y \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= 2 y \frac{1}{2} e^{u}

Substitute back

u = 2 z

= 2 y \frac{1}{2} e^{2 z}

Simplify

2 y \frac{1}{2} e^{2 z} : y e^{2 z}

= y e^{2 z}

Add a constant to the solution

= y e^{2 z} + C

h (g^{^{'}} (x))

y = \frac{4}{x}, y = 16 x, y = \frac{1}{16} x

y = 3 x ln (x)

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

f (x) = \frac{9}{x}, at x = 2