What is the integral of (3x^2-8x)/(x^3-4x^2+1) ?

The integral of (3x^2-8x)/(x^3-4x^2+1) is ln|x^3-4x^2+1|+C

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integral of (3x^2-8x)/(x^3-4x^2+1)

\int \frac{3 x ^{2} - 8 x}{x ^{3} - 4 x ^{2} + 1} d x

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

Solution steps

\int \frac{3 x ^{2} - 8 x}{x ^{3} - 4 x ^{2} + 1} 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^{2} + 1

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

Add a constant to the solution

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

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What is the integral of (3x^2-8x)/(x^3-4x^2+1) ?
The integral of (3x^2-8x)/(x^3-4x^2+1) is ln|x^3-4x^2+1|+C