What is the integral of e^{3x-1} ?

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

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integral of e^{3x-1}

integral

e^{3 x - 1}

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

Solution steps

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

Apply u-substitution

: \int e^{u} \frac{1}{3} d u

= \int e^{u} \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{3} e^{u}

Substitute back

u = 3 x - 1

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

Add a constant to the solution

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

e^{2}

e^{3 x - 2}

f (x) = (\frac{x}{x + 40})^{\frac{1}{2}}

\frac{x}{( x ^{2} + 6 ) 4}

x (lo g_{10} (x))^{2}