SOLUTION
Write out ythe expression
Let
[tex]u=x^3-5x-8[/tex]
Differrentiate u with respect to x
[tex]\begin{gathered} \frac{du}{dx}=3x^2-5 \\ \text{Then} \\ dx=\frac{du}{3x^2-5} \end{gathered}[/tex]
Then, substitute into the expression above
[tex]\int \frac{-3x^2+5}{x^3-5x-8}dx=\int \frac{-3x^2+5}{u}\times\frac{du}{3x^2-5}[/tex]
Then
[tex]\begin{gathered} \int \frac{-(3x^2-5)}{u}\times\frac{du}{3x^2-5} \\ \text{Divide the common factor } \\ \int -\frac{1}{u}du \end{gathered}[/tex]
Apply the rule
[tex]\begin{gathered} \int a\cdot f\mleft(x\mright)dx=a\cdot\int f\mleft(x\mright)dx \\ \text{Then} \\ \int -\frac{1}{u}du=-\int \frac{1}{u}du \end{gathered}[/tex]
Then use the common integral rule
[tex]\int \frac{1}{u}du=\ln \mleft(\mleft|u\mright|\mright)[/tex]
Replace the expression for u, we have
[tex]-\ln (|u|)=-\ln \mleft|x^3-5x-8\mright|+C[/tex]
Therefore
The solution becomes
- ln |x³-5x - 8 | +C
