What is the integral of (x^2)/(8x^3-1) ?

The integral of (x^2)/(8x^3-1) is 1/24 ln|8x^3-1|+C

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integral of (x^2)/(8x^3-1)

\int \frac{x ^{2}}{8 x ^{3} - 1} d x

\frac{1}{24} ln 8 x^{3} - 1 + C

Solution steps

\int \frac{x ^{2}}{8 x ^{3} - 1} d x

Apply u-substitution

: \int \frac{1}{24 u} d u

= \int \frac{1}{24 u} d u

Take the constant out:

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

= \frac{1}{24} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{24} ln ∣ u ∣

Substitute back

u = 8 x^{3} - 1

= \frac{1}{24} ln 8 x^{3} - 1

Add a constant to the solution

= \frac{1}{24} ln 8 x^{3} - 1 + C

(2.4) (1.6)

f (x) = \frac{2}{x - 1}, at x = 3

\int 2 sin^{2} (t) d t

\frac{d}{d x} (x^{3} + 4 x)

y^{^{'}} + 2 y = 4

What is the integral of (x^2)/(8x^3-1) ?
The integral of (x^2)/(8x^3-1) is 1/24 ln|8x^3-1|+C