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

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

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

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

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

Solution steps

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

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 = - t^{3} + 9 t + 1

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

Add a constant to the solution

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

3 x

\frac{d}{d x} (co x)

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

\frac{\partial}{\partial x} (8 x cos (5 x y))

x \to 0 + lim (\frac{0}{x})

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