It's been a while since I've done an Integral, but am required to relearn them for a class. Could anyone help me with the integral of $\dfrac{1}{i^2}$? Wouldn't the answer be $-1/i$ ?
Context
I'm trying to use the integral test for the summation $\sum_{i=1 }^n 1/i^2$. I'm trying to show that that summation is bounded above by a constant, so I'm using the integral test to do it. I'm just having trouble integrating $1/i^2$.
so the power rules is: $\int x^n dx=\frac{x^{n+1}}{n+1}$, which holds for every number $n$, with the exception of $n=-1$. In this case, you have $\int i^{-2} di$ Can you see how to proceed?