What is the integral of 1/((1-3x)) ?

The integral of 1/((1-3x)) is -1/3 ln|1-3x|+C

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integral of 1/((1-3x))

\int \frac{1}{( 1 - 3 x )} d x

- \frac{1}{3} ln ∣ 1 - 3 x ∣ + C

Solution steps

\int \frac{1}{1 - 3 x} d x

Apply u-substitution

= \int - \frac{1}{3 u} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 - 3 x

= - \frac{1}{3} ln ∣ 1 - 3 x ∣

Add a constant to the solution

= - \frac{1}{3} ln ∣ 1 - 3 x ∣ + C

\int x \cdot 4 x^{2} + 1 d x

\int a u d u

\int \frac{x - 7}{x ^{2} + 16} d x

\int \frac{7}{( x - 3 ) ^{2}} d x

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

What is the integral of 1/((1-3x)) ?
The integral of 1/((1-3x)) is -1/3 ln|1-3x|+C