What is the integral of x/(6x^2-18) ?

The integral of x/(6x^2-18) is 1/12 ln|6x^2-18|+C

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integral of x/(6x^2-18)

\int \frac{x}{6 x ^{2} - 18} d x

\frac{1}{12} ln 6 x^{2} - 18 + C

Solution steps

\int \frac{x}{6 x ^{2} - 18} d x

Apply u-substitution

: \int \frac{1}{12 u} d u

= \int \frac{1}{12 u} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{12} ln ∣ u ∣

Substitute back

u = 6 x^{2} - 18

= \frac{1}{12} ln 6 x^{2} - 18

Add a constant to the solution

= \frac{1}{12} ln 6 x^{2} - 18 + C

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What is the integral of x/(6x^2-18) ?
The integral of x/(6x^2-18) is 1/12 ln|6x^2-18|+C