What is the integral of 1/(9-20x) ?

The integral of 1/(9-20x) is -1/20 ln|9-20x|+C

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integral of 1/(9-20x)

\int \frac{1}{9 - 20 x} d x

- \frac{1}{20} ln ∣ 9 - 20 x ∣ + C

Solution steps

\int \frac{1}{9 - 20 x} d x

Apply u-substitution

: \int - \frac{1}{20 u} d u

= \int - \frac{1}{20 u} d u

Take the constant out:

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

= - \frac{1}{20} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{20} ln ∣ u ∣

Substitute back

u = 9 - 20 x

= - \frac{1}{20} ln ∣ 9 - 20 x ∣

Add a constant to the solution

= - \frac{1}{20} ln ∣ 9 - 20 x ∣ + C

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What is the integral of 1/(9-20x) ?
The integral of 1/(9-20x) is -1/20 ln|9-20x|+C