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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = x^{3} - 3 x + 1

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

Add a constant to the solution

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

cos^{s i n (x)} (x)

f (x) = e^{- 0.5 x} (4 - x) - 2

n = 1 \sum \infty \frac{5}{4 n ^{2} - 1}

\frac{d}{d x} (- 3 sin (6 x))

3 x, \frac{6 x}{7}, 55 - x^{2}

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