Let $0<t<1$. I want to show that $$ (1-t)\int_0^t\frac{1}{1-s}\,dB_s\overset{d}{=}B_t-tB_1. $$
I tried using Ito's formula for a standard process. That is for $f(t,x):=(1-t)x$, denoting $X_t:=\int_0^t\frac{1}{1-s}\,dB_s$, we have $f\in C^{1,2}(\mathbb{R}^+\times\mathbb{R})$, so $$ \begin{split} (1-t)X_t = f(t,X_t) &= f(0,0)+\int_0^t\frac{\partial f}{\partial x}(s,X_s)\,dX_s\\ &\qquad + \int_0^t\frac{\partial f}{\partial t}\left( s,X_s \right)\,ds + \frac{1}{2}\int_0^t\frac{\partial^2 f}{\partial x^2}(s,X_s)\frac{1}{(1-s)^2}\,ds\\ &= \int_0^t \frac{1-s}{1-s}\,dB_s-\int_0^t X_s\,ds\\ &= B_t - \int_0^tX_s\,ds. \end{split} $$ This is where I'm stuck. I think I need to show that the right-hand side has a normal distribution with mean $0$ and variance $t(1-t)$ but I don't know how to proceed. I'd appreciate any help.