What is the integral of x/(3x^2-1) ?

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

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integral of x/(3x^2-1)

\int \frac{x}{3 x ^{2} - 1} d x

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

Solution steps

\int \frac{x}{3 x ^{2} - 1} d x

Apply u-substitution

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

= \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} - 1

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

Add a constant to the solution

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

(- 10.9) (- 12.7)

sin (\frac{2 x}{x + 1})

\int 7 x^{2} y + \frac{3}{x} - 1 d x

\frac{\partial}{\partial x} (12 y^{2} x^{2} - 6 y x)

\frac{d}{d x} (sin^{4} (4 x))

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