What is the integral of 1/(7-6x) ?

The integral of 1/(7-6x) is -1/6 ln|7-6x|+C

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integral of 1/(7-6x)

\int \frac{1}{7 - 6 x} d x

- \frac{1}{6} ln ∣ 7 - 6 x ∣ + C

Solution steps

\int \frac{1}{7 - 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 = 7 - 6 x

= - \frac{1}{6} ln ∣ 7 - 6 x ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ 7 - 6 x ∣ + C

x \to 5 - lim (\frac{x + 1}{x - 5})

f (x) = - cos (x)

\frac{d}{d x} (x (2 x + 1)^{3})

f (x) = (4 x^{2} - 3 x)^{2}, at x = 1

\int \frac{x ^{2} + 20 x - 4}{x ^{3} - 4 x} d x