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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x^{3} - 4 x

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

Add a constant to the solution

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

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