What is the integral of (24x^3-6)/(2x^4-2x+3) ?

The integral of (24x^3-6)/(2x^4-2x+3) is 3ln|2x^4-2x+3|+C

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integral of (24x^3-6)/(2x^4-2x+3)

\int \frac{24 x ^{3} - 6}{2 x ^{4} - 2 x + 3} d x

3 ln 2 x^{4} - 2 x + 3 + C

Solution steps

\int \frac{24 x ^{3} - 6}{2 x ^{4} - 2 x + 3} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 ln ∣ u ∣

Substitute back

u = 2 x^{4} - 2 x + 3

= 3 ln 2 x^{4} - 2 x + 3

Add a constant to the solution

= 3 ln 2 x^{4} - 2 x + 3 + C

x \to 0 lim (5 - x^{2})

x \to 1 + lim (\frac{9}{x ^{3} - 1})

- 5 x^{2} + 3 y^{2} = 1

\frac{\partial}{\partial x} (x yz w + 2 x + 3 y + 4 z + w)

\frac{d}{d x} ((\frac{2 x}{x ^{2} + 1})^{2})

What is the integral of (24x^3-6)/(2x^4-2x+3) ?
The integral of (24x^3-6)/(2x^4-2x+3) is 3ln|2x^4-2x+3|+C