What is the integral of (6x^2)/(x^3-1) ?

The integral of (6x^2)/(x^3-1) is 2ln|x^3-1|+C

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integral of (6x^2)/(x^3-1)

\int \frac{6 x ^{2}}{x ^{3} - 1} d x

2 ln x^{3} - 1 + C

Solution steps

\int \frac{6 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

= 6 \cdot \int \frac{x ^{2}}{x ^{3} - 1} d x

Apply u-substitution

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

= 6 \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

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{3} ln ∣ u ∣

Substitute back

u = x^{3} - 1

= 6 \cdot \frac{1}{3} ln x^{3} - 1

Simplify

6 \cdot \frac{1}{3} ln x^{3} - 1 : 2 ln x^{3} - 1

= 2 ln x^{3} - 1

Add a constant to the solution

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

3 s

y = 7 x + 5

9 (x^{3} - x)^{4}

\frac{\partial}{\partial x} (5 sin (2 (5 x + 6 y)))

\int x^{2} \cdot x^{3} + 1 d x

What is the integral of (6x^2)/(x^3-1) ?
The integral of (6x^2)/(x^3-1) is 2ln|x^3-1|+C