I found the following integral
$\int_{0}^1 B_t t^{-1}dt,$
where $B_t$ is a standard Brownian motion. Using Ito formula with $f(t,x)=x\log(t)$ I achieved
$$0=\log(1)B_1=\int_{0}^1 B_s s^{-1}ds +\int_{0}^1 \log(s)dB_s,$$
and further by properties of the stochastic integral $\int_{0}^1 \log(s)dB_s\sim N(0,\sigma^2)$, where
$\sigma^2=\int_0^1\log^2(s)ds=2$
So the above r.v. $\int_{0}^1 B_t t^{-1}dt$ is $N(0,2)$ distributed. Is this right? Is there any problem concerning the definition of $\int_{0}^1 B_t t^{-1}dt$? I mean $B_t t^{-1}\sim N(0,t^{-1})$ and the variance goes to infinity.
You seen to have done everything formally right. However you're basically applying the stochastic integration by parts rule (oksendal p. 46), which requires the integrand of the ito integral (here $log(s)$) to be continuous ( hence bounded) on [0,1]. I don't know for certain that your result is wrong, but one of the conditions of the relevant theorem isn't satisfied.
EDIT: actually the way to test this would be to try to compute the second moment of the first integral directly, without applying integration by parts. You can do that applying the formula in my answer here: Probability density function of the integral of a continuous stochastic process
I suspect it will diverge, which would indicate that your integral is not well defined. But maybe I'm wrong.