What is the integral of (x^3-3x)/(x^4-6x^2-9) ?

The integral of (x^3-3x)/(x^4-6x^2-9) is 1/4 ln|x^4-6x^2-9|+C

integral of (x^3-3x)/(x^4-6x^2-9)

\int \frac{x ^{3} - 3 x}{x ^{4} - 6 x ^{2} - 9} d x

\frac{1}{4} ln x^{4} - 6 x^{2} - 9 + C

Solution steps

\int \frac{x ^{3} - 3 x}{x ^{4} - 6 x ^{2} - 9} d x

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = x^{4} - 6 x^{2} - 9

= \frac{1}{4} ln x^{4} - 6 x^{2} - 9

Add a constant to the solution

= \frac{1}{4} ln x^{4} - 6 x^{2} - 9 + C

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

\int e^{5 x} 5 d x

\int \frac{x ^{2} + 5}{x} d x

\int \frac{1}{5 3 x ^{5}} d x

\int \frac{x ^{\frac{1}{2}}}{x + x ^{\frac{1}{4}}} d x

What is the integral of (x^3-3x)/(x^4-6x^2-9) ?
The integral of (x^3-3x)/(x^4-6x^2-9) is 1/4 ln|x^4-6x^2-9|+C