Question

How do you find the indefinite integral of `int ((x²) / (16-x³)²) dx`?

Answer 1

`(-1/3)*ln(16-x^3)+c`

Explanation:

Don't know if the 2 after the fraction and before the closing bracket is a scalar multiple or an error, so doing it without the 2.
replace `16-x^3` with `t`
so we have `-3*x^2dx = dt`
the equation becomes:
`int(-1/(3*t))dt` = `(-1/3)*ln(t)+c`
now replacing t back,
`(-1/3)*ln(16-x^3)+c`