Why the integral $\int_0^3 \frac{1}{x^2-6x+5}dx$ doesn't exist? Why the integral $\int_0^3 \frac{1}{x^2-6x+5}dx$ doesn't exist?
$$
I=\frac{1}{2} \lim_{t\to 1^{-}} [\arctan(2-3/2) - \arctan(-3/2)] + \lim_{t\to 1^+}[\arctan(0/2) - \arctan(1-3/2)]
$$
I can find $\arctan(0)= \pi/2$
So why the answer in my book is doesn't exist?
 A: Note that $\frac{1}{(x-1)(x-5)}$ is undefined at $x = 1$ so 
$$\int_0^3\frac{1}{(x - 1)(x-5)}dx = \lim_{s\to 1^-}\int_0^s\frac{1}{(x-1)(x-5)}dx + \lim_{t\to 1^+}\int_t^3\frac{1}{(x-1)(x-5)}dx.$$
To evaluate the two integrals on the right, we need to find an antiderivative. We do this by first applying partial fraction. Doing so we obtain
$$\frac{1}{(x-1)(x-5)} = \frac{-\frac{1}{4}}{x-1} +\frac{\frac{1}{4}}{x-5}.$$
Therefore $-\frac{1}{4}\ln|x-1| + \frac{1}{4}\ln|x-5| = \frac{1}{4}\ln\left|\frac{x-5}{x-1}\right|$ is an antiderivative of $\frac{1}{(x-1)(x-5)}$. Therefore
$$\int_0^s\frac{1}{(x-1)(x-5)}dx = \frac{1}{4}\ln\left|\frac{s-5}{s-1}\right| - \frac{1}{4}\ln\left|\frac{0-5}{0-1}\right| = \frac{1}{4}\ln\left|\frac{s-5}{s-1}\right| -\frac{1}{4}\ln 5.$$
As $\lim\limits_{s\to 1^-}\frac{1}{4}\ln\left|\frac{s-5}{s-1}\right|$ does not exist, $\lim\limits_{s\to 1^-}\int_0^s\frac{1}{(x-1)(x-5)}dx$ does not exist, so $\int_0^3\frac{1}{(x-1)(x-5)}dx$ does not exist.
A: The improper integral: $$\int_1^3f(x)dx$$ does not converge. In fact, using Quotient test for it gives us: $$\lim_{x\to 1^+}(x-1)^{\color{red}{1}}f(x)\neq 0$$ and so it diverges. 
A: 
Let $a,b,\alpha$ be real numbers such that $a<b$.
The integrals $\displaystyle \int \limits_a^{b^-} \frac{1}{(b-x)^\alpha}\mathrm dx , \int \limits_{a^+}^b \frac{1}{(x-a)^\alpha}\mathrm dx$ converge if, and only if, $\alpha <1$.

Hint: $\forall x\in \mathbb R\left((x^2-6x+5)=(x-1)(x-5)\right)$.
A: 
This is a plot of the function $y=\frac{1}{x^2-6x+5}$. Does it look integrable to you on the interval $[0,3]$? (Hint: pay special attention to the behaviour of the function around $x=1$.)
