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

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

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

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

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

Solution steps

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

= ln x^{3} - x + 1

Add a constant to the solution

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

\int sin (x) ln (1 + sin (x)) d x

\int e^{- ik x} d x

\int \frac{1}{e ^{2 x} - 9} d x

\int \frac{1}{x} \frac{x + 1}{x} d x

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