This improper integral doesn't converge, does it?

Solution 1:

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.

Solution 2:

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: $$ \int_1^3 \frac{1}{x-3}dx = \lim_{b\to 3}\int_1^b\frac{1}{x-3} dx = \lim_{b\to 3}\left[\log|x-3|\right]_1^b = \infty. $$