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

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

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integral of 1/(4x-1)

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

\frac{1}{4} ln ∣ 4 x - 1 ∣ + C

Solution steps

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

Apply u-substitution

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

= \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 - 1

= \frac{1}{4} ln ∣ 4 x - 1 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ 4 x - 1 ∣ + C

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y^{^{''}} + 3 y = sin (t)

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