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

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

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integral of 1/(4x-3)

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

\frac{1}{4} ln ∣ 4 x - 3 ∣ + C

Solution steps

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

Apply u-substitution

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = 4 x - 3

= \frac{1}{4} ln ∣ 4 x - 3 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ 4 x - 3 ∣ + C