What is the integral of 1/(1-8x) ?

The integral of 1/(1-8x) is -1/8 ln|1-8x|+C

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integral of 1/(1-8x)

\int \frac{1}{1 - 8 x} d x

- \frac{1}{8} ln ∣ 1 - 8 x ∣ + C

Solution steps

\int \frac{1}{1 - 8 x} d x

Apply u-substitution

: \int - \frac{1}{8 u} d u

= \int - \frac{1}{8 u} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{8} ln ∣ u ∣

Substitute back

u = 1 - 8 x

= - \frac{1}{8} ln ∣ 1 - 8 x ∣

Add a constant to the solution

= - \frac{1}{8} ln ∣ 1 - 8 x ∣ + C

f (x) = \frac{9 x}{( x + 1 ) ^{2}}, (0, 0)

\frac{1}{s ( s ^{2} + 4 )}

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\frac{d y}{d x} = x x - 15