What is the integral of e^{6x-3} ?

The integral of e^{6x-3} is 1/6 e^{6x-3}+C

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integral of e^{6x-3}

\int e^{6 x - 3} d x

\frac{1}{6} e^{6 x - 3} + C

Solution steps

\int e^{6 x - 3} d x

Apply u-substitution

: \int e^{u} \frac{1}{6} d u

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{6} e^{u}

Substitute back

u = 6 x - 3

= \frac{1}{6} e^{6 x - 3}

Add a constant to the solution

= \frac{1}{6} e^{6 x - 3} + C

f (x) = x (36 + x^{2})^{- \frac{1}{2}}, at x = 0

113 t - 0.5 t^{4} + 900

\frac{d}{d x} (\frac{1}{x e ^{x}})

\frac{\partial}{\partial s} (3 s + t)

f (x) = 3 csc (x) - 5 sec (x)