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

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

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integral of 1/(4xln(x)-3x)

\int \frac{1}{4 x ln ( x ) - 3 x} d x

\frac{1}{4} ln ∣ 4 ln (x) - 3 ∣ + C

Solution steps

\int \frac{1}{4 x ln ( x ) - 3 x} d x

Apply u-substitution

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

Apply u-substitution

= \int \frac{1}{4 v} d v

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}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{4} ln ∣ v ∣

Substitute back

= \frac{1}{4} ln ∣ 4 ln (x) - 3 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ 4 ln (x) - 3 ∣ + C

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