What is the integral of 9e^{5x+e^{5x}} ?

The integral of 9e^{5x+e^{5x}} is 9/5 e^{e^{5x}}+C

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integral of 9e^{5x+e^{5x}}

\int 9 e^{5 x + e^{5 x}} d x

\frac{9}{5} e^{e^{5 x}} + C

Solution steps

\int 9 e^{5 x + e^{5 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^{5 x + e^{5 x}} d x

Apply u-substitution

: \int e^{u + e^{u}} \frac{1}{5} d u

= 9 \cdot \int e^{u + e^{u}} \frac{1}{5} d u

Take the constant out:

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

= 9 \cdot \frac{1}{5} \cdot \int e^{u + e^{u}} d u

Apply u-substitution

: \int e^{v} d v

= 9 \cdot \frac{1}{5} \cdot \int e^{v} d v

Use the common integral:

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

= 9 \cdot \frac{1}{5} e^{v}

Substitute back

= 9 \cdot \frac{1}{5} e^{e^{5 x}}

Simplify

9 \cdot \frac{1}{5} e^{e^{5 x}} : \frac{9}{5} e^{e^{5 x}}

= \frac{9}{5} e^{e^{5 x}}

Add a constant to the solution

= \frac{9}{5} e^{e^{5 x}} + C

\int x (4 x - 1)^{4} d x

\int \frac{1}{4 + ( u + 3 )} d u

y = ln (8 x)

y^{^{''}} - 7 y^{^{'}} = 5 t

x \to 0 lim (\frac{x}{sin ( 4 x )})

What is the integral of 9e^{5x+e^{5x}} ?
The integral of 9e^{5x+e^{5x}} is 9/5 e^{e^{5x}}+C