`int(t^2-3)/(-t^3+9t+1)dt`
Let's apply integral substitution,
Substitute `u=(-t^3+9t+1)`
`(du)/dt=-3t^2+9`
`(du)/dt=-3(t^2-3)`
`=>-1/3(du)=(t^2-3)dt`
`int(t2-3)/(-t^3+9t+1)dt=int(-1/3)(du)/u`
Take the constant out,
`=-1/3int(du)/u`
use the common integral:`int1/udu=ln|u|`
`=-1/3ln|u|`
substitute back `u=(-t^3+9t+1)`
and add a constant C to the solution,
`=-1/3ln|(-t^3+9t+1)|+C`
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