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

The integral of e^{4x+e^{4x}} is 1/4 e^{e^{4x}}+C

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integral of e^{4x+e^{4x}}

\int e^{4 x + e^{4 x}} d x

\frac{1}{4} e^{e^{4 x}} + C

Solution steps

\int e^{4 x + e^{4 x}} d x

Apply u-substitution

: \int e^{u + e^{u}} \frac{1}{4} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int e^{v} d v

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

Use the common integral:

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

= \frac{1}{4} e^{v}

Substitute back

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

Add a constant to the solution

= \frac{1}{4} e^{e^{4 x}} + C

y^{^{'}} + \frac{1}{20 x} y = - 9.81

\frac{d}{d x} (5 - cos (x))

\frac{\partial}{\partial x} (2 (x - 1)^{2} - 3 y - 5)

y^{^{'}} = x^{3} (1 - y)

y^{^{'}} = 0.001 y (1800 - y)

What is the integral of e^{4x+e^{4x}} ?
The integral of e^{4x+e^{4x}} is 1/4 e^{e^{4x}}+C