What is the integral of (x^2-3)/(-x^3+9x+1) ?

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

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integral of (x^2-3)/(-x^3+9x+1)

\int \frac{x ^{2} - 3}{- x ^{3} + 9 x + 1} d x

- \frac{1}{3} ln - x^{3} + 9 x + 1 + C

Solution steps

\int \frac{x ^{2} - 3}{- x ^{3} + 9 x + 1} 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} + 9 x + 1

= - \frac{1}{3} ln - x^{3} + 9 x + 1

Add a constant to the solution

= - \frac{1}{3} ln - x^{3} + 9 x + 1 + C

\int (2 x - 3) (5 x + 1) d x

\int_{1}^{5} \frac{4}{x} d x

n = 1 \sum \infty \frac{2}{3 ^{n} + 1}

\frac{d}{d x} (cot (tan (x)))

y^{^{'}} = (1 - 2 x) y^{2}, y (0) = - \frac{1}{30}

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