What is the integral of e^{6u} ?

The integral of e^{6u} is 1/6 e^{6u}+C

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integral of e^{6u}

\int e^{6 u} d u

\frac{1}{6} e^{6 u} + C

Solution steps

\int e^{6 u} d u

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

v = 6 u

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

Add a constant to the solution

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

f (x) = \frac{x - 9}{8 x - 1}, at x = 1

lo g_{3} (x)

y^{^{'''}} + 6 y^{^{''}} - 15 y^{^{'}} + 8 y = 0

\frac{d}{d x} (\frac{x ^{2}}{9 + 4 x})

\int_{\frac{1}{7}}^{5} 8 x ln (7 x) d x