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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 2 \cdot \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

= 2 \cdot \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{6} ln ∣ u ∣

Substitute back

u = 3 x^{2} - 4

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

Simplify

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

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

Add a constant to the solution

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

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