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

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

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

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

ln x^{3} - 4 x + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x^{3} - 4 x

= ln x^{3} - 4 x

Add a constant to the solution

= ln x^{3} - 4 x + C

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\int x^{2} (x^{3} - 2) d x

\int \frac{400 x ^{2}}{x ^{3} + 4600} d x

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