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

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

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

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

ln x^{3} - 1 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{3} ln ∣ u ∣

Substitute back

u = x^{3} - 1

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

Simplify

3 \cdot \frac{1}{3} ln x^{3} - 1 : ln x^{3} - 1

= ln x^{3} - 1

Add a constant to the solution

= ln x^{3} - 1 + C

f (x) = (ln (x^{5}))^{2}

\frac{d}{d x} (- \frac{1}{2} x cos (x))

\int \frac{9}{x ^{- 1} + 3} d x

\frac{\partial}{\partial x} (y cos (x^{2} y))

(8 \frac{1}{7}, 9), (4, 3 \frac{1}{5})

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