What is the integral of (3x)/(2x^2-6) ?

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

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integral of (3x)/(2x^2-6)

\int \frac{3 x}{2 x ^{2} - 6} d x

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

Solution steps

\int \frac{3 x}{2 x ^{2} - 6} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{2 x ^{2} - 6} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = 2 x^{2} - 6

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

Simplify

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

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

Add a constant to the solution

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

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