What is the integral of 9e^{4x+e^{4x}} ?

The integral of 9e^{4x+e^{4x}} is 9/4 e^{e^{4x}}+C

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integral of 9e^{4x+e^{4x}}

\int 9 e^{4 x + e^{4 x}} d x

\frac{9}{4} e^{e^{4 x}} + C

Solution steps

\int 9 e^{4 x + e^{4 x}} d x

Take the constant out:

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

= 9 \cdot \int e^{4 x + e^{4 x}} d x

Apply u-substitution

= 9 \cdot \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

= 9 \cdot \frac{1}{4} \cdot \int e^{u + e^{u}} d u

Apply u-substitution

= 9 \cdot \frac{1}{4} \cdot \int e^{v} d v

Use the common integral:

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

= 9 \cdot \frac{1}{4} e^{v}

Substitute back

= 9 \cdot \frac{1}{4} e^{e^{4 x}}

Simplify

9 \cdot \frac{1}{4} e^{e^{4 x}} : \frac{9}{4} e^{e^{4 x}}

= \frac{9}{4} e^{e^{4 x}}

Add a constant to the solution

= \frac{9}{4} e^{e^{4 x}} + C

What is the integral of 9e^{4x+e^{4x}} ?
The integral of 9e^{4x+e^{4x}} is 9/4 e^{e^{4x}}+C