What is the integral of 5e^{4x+e^{4x}} ?

The integral of 5e^{4x+e^{4x}} is 5/4 e^{e^{4x}}+C

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integral of 5e^{4x+e^{4x}}

\int 5 e^{4 x + e^{4 x}} d x

\frac{5}{4} e^{e^{4 x}} + C

Solution steps

\int 5 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

= 5 \cdot \int e^{4 x + e^{4 x}} d x

Apply u-substitution

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

= 5 \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

= 5 \cdot \frac{1}{4} \cdot \int e^{u + e^{u}} d u

Apply u-substitution

: \int e^{v} d v

= 5 \cdot \frac{1}{4} \cdot \int e^{v} d v

Use the common integral:

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

= 5 \cdot \frac{1}{4} e^{v}

Substitute back

= 5 \cdot \frac{1}{4} e^{e^{4 x}}

Simplify

5 \cdot \frac{1}{4} e^{e^{4 x}} : \frac{5}{4} e^{e^{4 x}}

= \frac{5}{4} e^{e^{4 x}}

Add a constant to the solution

= \frac{5}{4} e^{e^{4 x}} + C

f (x) = (x^{3} + 3 x + 6) (4 + \frac{1}{x ^{2}})

\int cos (\frac{x}{4}) sin (\frac{x}{4}) d x

arctan (x^{3})

\int \frac{x + 3}{x ^{2} - 6 x + 5} d x

\frac{d}{d x} (ln (5 x + 7))

What is the integral of 5e^{4x+e^{4x}} ?
The integral of 5e^{4x+e^{4x}} is 5/4 e^{e^{4x}}+C