What is the integral of (12x^2)/(4x^3-6) ?

The integral of (12x^2)/(4x^3-6) is ln|4x^3-6|+C

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integral of (12x^2)/(4x^3-6)

\int \frac{12 x ^{2}}{4 x ^{3} - 6} d x

ln 4 x^{3} - 6 + C

Solution steps

\int \frac{12 x ^{2}}{4 x ^{3} - 6} d x

Take the constant out:

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

= 12 \cdot \int \frac{x ^{2}}{4 x ^{3} - 6} d x

Apply u-substitution

: \int \frac{1}{12 u} d u

= 12 \cdot \int \frac{1}{12 u} d u

Take the constant out:

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

= 12 \cdot \frac{1}{12} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 12 \cdot \frac{1}{12} ln ∣ u ∣

Substitute back

u = 4 x^{3} - 6

= 12 \cdot \frac{1}{12} ln 4 x^{3} - 6

Simplify

12 \cdot \frac{1}{12} ln 4 x^{3} - 6 : ln 4 x^{3} - 6

= ln 4 x^{3} - 6

Add a constant to the solution

= ln 4 x^{3} - 6 + C

y^{^{'}} = \frac{( 3 x sec ( \frac{y}{x} ) + y )}{x}

\frac{\partial}{\partial x} (\frac{x ^{2} - 2 y}{4 x ^{2} + 2 y})

\frac{\partial}{\partial y} (x y^{4})

f (x) = x^{3} - 3 x + 2, at x = 4

y = \frac{1}{x}, y = 2 x - 1, 0, 3

What is the integral of (12x^2)/(4x^3-6) ?
The integral of (12x^2)/(4x^3-6) is ln|4x^3-6|+C