What is the integral of x/((3x^2-4)) ?

The integral of x/((3x^2-4)) is 1/6 ln|3x^2-4|+C

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integral of x/((3x^2-4))

\int \frac{x}{( 3 x ^{2} - 4 )} d x

\frac{1}{6} ln 3 x^{2} - 4 + C

Solution steps

\int \frac{x}{3 x ^{2} - 4} d x

Apply u-substitution

= \int \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = 3 x^{2} - 4

= \frac{1}{6} ln 3 x^{2} - 4

Add a constant to the solution

= \frac{1}{6} ln 3 x^{2} - 4 + C

What is the integral of x/((3x^2-4)) ?
The integral of x/((3x^2-4)) is 1/6 ln|3x^2-4|+C