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

1 Answer

(-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