What is the integral of 1/7 e^{7x} ?

The integral of 1/7 e^{7x} is 1/49 e^{7x}+C

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integral of 1/7 e^{7x}

\int \frac{1}{7} e^{7 x} d x

\frac{1}{49} e^{7 x} + C

Solution steps

\int \frac{1}{7} e^{7 x} d x

Take the constant out:

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

= \frac{1}{7} \cdot \int e^{7 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{7} d u

= \frac{1}{7} \cdot \int e^{u} \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \frac{1}{7} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{7} \cdot \frac{1}{7} e^{u}

Substitute back

u = 7 x

= \frac{1}{7} \cdot \frac{1}{7} e^{7 x}

Simplify

\frac{1}{7} \cdot \frac{1}{7} e^{7 x} : \frac{1}{49} e^{7 x}

= \frac{1}{49} e^{7 x}

Add a constant to the solution

= \frac{1}{49} e^{7 x} + C

\frac{e ^{- 2 s}}{s ^{2} - 3 s - 4}

\frac{d}{d x} (ln (11))

x \to 0 lim (1 - \frac{1}{x})

(x - \frac{6}{x}), at m = 0

x^{8} e^{7 x}

What is the integral of 1/7 e^{7x} ?
The integral of 1/7 e^{7x} is 1/49 e^{7x}+C