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

The integral of (x^2-4)/(x^3-12x+3) is 1/3 ln|x^3-12x+3|+C

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

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

\frac{1}{3} ln x^{3} - 12 x + 3 + C

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = x^{3} - 12 x + 3

= \frac{1}{3} ln x^{3} - 12 x + 3

Add a constant to the solution

= \frac{1}{3} ln x^{3} - 12 x + 3 + C

\int 7^{x + 2} d x

\frac{d}{d x} (\frac{x ^{2} - 100}{x - 10})

- 2 e^{x}

\int (e^{x} + cos (x) + t + \frac{1}{t}) d x

\frac{\partial}{\partial x} (\frac{3 x}{2 + y})

What is the integral of (x^2-4)/(x^3-12x+3) ?
The integral of (x^2-4)/(x^3-12x+3) is 1/3 ln|x^3-12x+3|+C