What is the integral of e^{4x-1} ?

The integral of e^{4x-1} is 1/4 e^{4x-1}+C

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integral of e^{4x-1}

\int e^{4 x - 1} d x

\frac{1}{4} e^{4 x - 1} + C

Solution steps

\int e^{4 x - 1} d x

Apply u-substitution

= \int 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

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

Use the common integral:

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

= \frac{1}{4} e^{u}

Substitute back

u = 4 x - 1

= \frac{1}{4} e^{4 x - 1}

Add a constant to the solution

= \frac{1}{4} e^{4 x - 1} + C