What is the integral of 6e^{5x+e^{5x}} ?

The integral of 6e^{5x+e^{5x}} is 6/5 e^{e^{5x}}+C

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integral of 6e^{5x+e^{5x}}

\int 6 e^{5 x + e^{5 x}} d x

\frac{6}{5} e^{e^{5 x}} + C

Solution steps

\int 6 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

= 6 \cdot \int e^{5 x + e^{5 x}} d x

Apply u-substitution

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

= 6 \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

= 6 \cdot \frac{1}{5} \cdot \int e^{u + e^{u}} d u

Apply u-substitution

: \int e^{v} d v

= 6 \cdot \frac{1}{5} \cdot \int e^{v} d v

Use the common integral:

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

= 6 \cdot \frac{1}{5} e^{v}

Substitute back

= 6 \cdot \frac{1}{5} e^{e^{5 x}}

Simplify

6 \cdot \frac{1}{5} e^{e^{5 x}} : \frac{6}{5} e^{e^{5 x}}

= \frac{6}{5} e^{e^{5 x}}

Add a constant to the solution

= \frac{6}{5} e^{e^{5 x}} + C

\frac{d}{d y} (- \frac{x}{y})

x \to \infty lim (a x^{2})

\int \frac{3 x}{e ^{x}} d x

\frac{d}{d x} (x^{2} sin (\frac{1}{a x}))

x \to 1 - lim (\frac{2}{x ^{2} - 1})

What is the integral of 6e^{5x+e^{5x}} ?
The integral of 6e^{5x+e^{5x}} is 6/5 e^{e^{5x}}+C