1∫^2 1/(x-1)^2dx.

Answer 1

This is an improper integral, sometimes referred to as a type 2 improper integral (due to the vertical asymptote at x = 1). To deal with this we should consider the limit of proper integrals over intervals starting just the right of x = 1.

That is ∫₁²(x - 1)^-2dx = lim_{a -> 1} ∫_a²(x - 1)^-2dx = lim_{a -> 1} -(x-1)^-1 |_a² = lim_{a -> 1} -1 + (a - 1)^-1 = ∞.

It is hard to convey with this tool, but all above limits should be taken as a approaching 1 from the right, which is why the limit ultimately exists (the 2-sided limit of (a - 1)^-1 does not exist). Also note that it is a convention to allow infinite limits when appropriate and I know some instructors would write the above answer as "does not exist" since they don't want to allow for infinite limits. You should adjust accordingly based on your instructor's preference.

The best and most specific answer is ∞ with the supporting work I've written above (add plus sign superscripts to the 1 in the limit to denote right handed limits). "DNE" is also acceptable and perhaps preferred by some instructors. Most likely the important thing to get most credit is to show your work by inserting the limit.

Answer 2

Can't put 0 in a denominator (1-1)^2.