What is the integral of (e^{7x-3})/1 ?

The integral of (e^{7x-3})/1 is 1/7 e^{7x-3}+C

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integral of (e^{7x-3})/1

\int \frac{e ^{7 x - 3}}{1} d x

\frac{1}{7} e^{7 x - 3} + C

Solution steps

\int \frac{e ^{7 x - 3}}{1} d x

Simplify

= \int e^{7 x - 3} d x

Apply u-substitution

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

= \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 \int e^{u} d u

Use the common integral:

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

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

Substitute back

u = 7 x - 3

= \frac{1}{7} e^{7 x - 3}

Add a constant to the solution

= \frac{1}{7} e^{7 x - 3} + C

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What is the integral of (e^{7x-3})/1 ?
The integral of (e^{7x-3})/1 is 1/7 e^{7x-3}+C