Strange AP Calculus BC question help? The question is $\int_0^3\frac{1}{(1-x)^2}$. I got an answer (from u-substitution) however the solution manual says that the integral does not converge. Someone told me that the integral is undefined at $x = 1$, but if we look at $\frac {1}{x}$ that function is undefined at $x=0$ yet we can still integrate it. 
Can someone explain what is going on?! 

As was pointed out to you, the integrand fails to exist at $x=1$, so the integral must be rewritten as the sum of two improper integrals, one from $0$ to $1$ and the other from $1$ to $3$. When you try to evaluate the latter, you get this:
$$\begin{align*}\int_1^3\frac{dx}{(1-x)^2}&=\lim_{a\to 1^+}\int_a^3\frac{dx}{(1-x)^2}\\\\
&=\lim_{a\to 1^+}\left[\frac1{1-x}\right]_a^3\\\\
&=-\frac12-\lim_{a\to 1^+}\frac1{1-a}\;,
\end{align*}$$
and the limit in the last line clearly does not exist. Thus, the improper integral diverges, as does your original integral.
Note that you cannot integrate the function $f(x)=\frac1x$ over any interval that contains $0$, and the same goes for the function $f(x)=\frac1{x^2}$: in both cases you will get a divergent improper integral.
