What is the integral of 3/(1-4x) ?

The integral of 3/(1-4x) is -3/4 ln|1-4x|+C

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integral of 3/(1-4x)

\int \frac{3}{1 - 4 x} d x

- \frac{3}{4} ln ∣ 1 - 4 x ∣ + C

Solution steps

\int \frac{3}{1 - 4 x} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{1 - 4 x} d x

Apply u-substitution

: \int - \frac{1}{4 u} d u

= 3 \cdot \int - \frac{1}{4 u} d u

Take the constant out:

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

= 3 (- \frac{1}{4} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 (- \frac{1}{4} ln ∣ u ∣)

Substitute back

u = 1 - 4 x

= 3 (- \frac{1}{4} ln ∣ 1 - 4 x ∣)

Simplify

3 (- \frac{1}{4} ln ∣ 1 - 4 x ∣) : - \frac{3}{4} ln ∣ 1 - 4 x ∣

= - \frac{3}{4} ln ∣ 1 - 4 x ∣

Add a constant to the solution

= - \frac{3}{4} ln ∣ 1 - 4 x ∣ + C

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