What is the integral of 1/(9-6x) ?

The integral of 1/(9-6x) is -1/6 ln|9-6x|+C

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integral of 1/(9-6x)

\int \frac{1}{9 - 6 x} d x

- \frac{1}{6} ln ∣ 9 - 6 x ∣ + C

Solution steps

\int \frac{1}{9 - 6 x} d x

Apply u-substitution

: \int - \frac{1}{6 u} d u

= \int - \frac{1}{6 u} 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 \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{6} ln ∣ u ∣

Substitute back

u = 9 - 6 x

= - \frac{1}{6} ln ∣ 9 - 6 x ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ 9 - 6 x ∣ + C

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

\int sin (\frac{x}{2}) d x

f (x) = \frac{1}{cos ( x )}

\frac{d}{d y} (2 x y^{4} e^{y} + 2 x y^{3} + y)

y^{^{''}} + 4 y^{^{'}} + 6 y = cos (x), y (0) = 0, y^{^{'}} (0) = 2