This improper integral doesn't converge, does it? (disclaimer: I don't know if it's legit to share a picture of my calculations, and not rewriting it in LATEX notation - let me know if this is legit or not)
my question is that: are my calculations correct?
I have to solve this exercise: it's an improper integral.
before using integration by parts, I've studied the bounds, in order to check where the function is undefined. And I got that when solving the function for $x = 3$ the denominator is 0, therefore the function is undefined (i.e division by zero).
Therefore, I've taken the limit as x approaches 3. And then I've solved the integral with $x$ as the upper bound.
As stated above, I've used integration by parts by choosing $$ u = x $$ (because the derivative of the polynomial, hopefully, is going to become some smaller value) And $$ dv = (1 / (3 - x)) $$ because the antiderivative is simply equal to log (natural log, i.e with base $e$) (of course the argument of log must be the absolute value)
Now, by integrating by parts and after having evaluated the limit of the antiderivative, I got that the limit doesn't exist, because the limit of the function evaluated in the upper bound is undefined (i.e the natural log is undefined for x = 3). Is it true? And if it's true, the integral doesn't converge, right?

 A: You are correct that the integral does not converge, but you made some mistakes and overcomplicated the solution in general.
The mistake: If $v=\log(3-x)$, then $v'=-\frac{1}{3-x}$. You missed a minus sign.
The overcomplication:
Instead of using per partes, you can rewrite
$$\frac{x}{3-x} = \frac{x-3+3}{3-x} = \frac{-(3-x)}{3-x} + \frac{3}{3-x} = \frac{3}{3-x} - 1$$
and only integrate after this rearrangement. No need for per partes, a simple introduction of a new variable $u=3-x$ is sufficient and you get (since $du = -dx$):
$$\int_1^3\frac{x}{3-x}dx = 3\int_1^3 \frac{1}{3-x}dx - \int_1^3 1dx = 3\int_2^0-\frac{1}{u}du - 2 = 3\int_0^2\frac1udu - 2$$
now you can either remember that the integral of $\frac{1}{u}$ diverges around $0$, or you can write it out, since
$$\int_0^2\frac1udu=\lim_{x\to 0}\int_x^2\frac1udu = \lim_{x\to 0} (\ln(2)-\ln(x))$$ and the limit above does not exist.
A: Since the objective is to decide on the convergence of the integral, you can make it a little simpler. In fact, since $g(x)=x$ is continuous on $[1,3]$ and $g(3)\ne 0$, the convergence of the original integral is equivalent to the convergence of $\int_1^3 \frac{1}{x-3}\,dx$. This last integral is divergent:
$$
\int_1^3 \frac{1}{x-3}dx = \lim_{b\to 3}\int_1^b\frac{1}{x-3} dx = \lim_{b\to 3}\left[\log|x-3|\right]_1^b = \infty.
$$
