What is the integral of (4x^2)/((8x^3-4)) ?

The integral of (4x^2)/((8x^3-4)) is 1/6 ln|8x^3-4|+C

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integral of (4x^2)/((8x^3-4))

\int \frac{4 x ^{2}}{( 8 x ^{3} - 4 )} d x

\frac{1}{6} ln 8 x^{3} - 4 + C

Solution steps

\int \frac{4 x ^{2}}{8 x ^{3} - 4} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{2}}{8 x ^{3} - 4} d x

Apply u-substitution

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

= 4 \cdot \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

= 4 \cdot \frac{1}{24} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{24} ln ∣ u ∣

Substitute back

u = 8 x^{3} - 4

= 4 \cdot \frac{1}{24} ln 8 x^{3} - 4

Simplify

4 \cdot \frac{1}{24} ln 8 x^{3} - 4 : \frac{1}{6} ln 8 x^{3} - 4

= \frac{1}{6} ln 8 x^{3} - 4

Add a constant to the solution

= \frac{1}{6} ln 8 x^{3} - 4 + C

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What is the integral of (4x^2)/((8x^3-4)) ?
The integral of (4x^2)/((8x^3-4)) is 1/6 ln|8x^3-4|+C