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?! 
 A: 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.
A: You have an improper integral due to your function being unbounded on the interval. Anyways, you can interpret it in terms of limits of Riemann integrals as follows:$$\int_0^3\frac1{(1-x)^2}\,\mathrm{d}x=\lim_{a\to1^-}\int_0^a\frac1{(1-x)^2}+\lim_{b\to1^+}\int_b^3\frac1{(1-x)^2}\,\mathrm{d}x$$
You can determine the indefinite integral of $1/(1-x)^2$ with ease:$$\int\frac{\mathrm{d}x}{(\underbrace{1-x}_{\text{u}})^2}=-\int\frac{\mathrm{d}u}{u^2}=\frac1{1-x}$$
Using the fundamental theorem of calculus we may then rewrite our definite integrals as follows:$$\int_0^a\frac{\mathrm{d}x}{(1-x)^2}=\frac1{1-a}-1\\\int_b^3\frac{\mathrm{d}x}{(1-x)^2}=-\frac12-\frac1{1-b}$$Recognize that in the limit as $a\to1^+$ and $b\to1^-$ both of these are unbounded, leaving us with a divergent integral.

Alternatively, recognize $\dfrac1{1-x}=1+x+x^2+\dots$; it is clear this series diverges for $x>1$. You could use this to reason this reason about the behavior of our integral and understand why it diverges (though it is not generally enough to show that it does indeed diverge).
