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

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

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

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

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

Solution steps

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

Apply u-substitution

= \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} - 2 x^{2} + x - 2

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

Add a constant to the solution

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

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