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

\int_{1}^{5} \frac{1}{x - 2} d x

n = 1 \sum \infty \frac{n}{1 + 3 n}

n = 1 \sum \infty \frac{1}{n ^{0.5}}

\frac{d}{d t} (4 e^{- 3 t})

\frac{d}{d A} (- cos (2) (A) - sin (2) (A))

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