so we know in general $\int_0^T f(s) dBs = f(t)B_t - \int_0^{T} f'(s) B_s ds$
what could I do with the second term $\int_0^{T} f'(s) B_s ds$? I have read somewhere that
$\int_0^{1} f'(s) B_s ds$ = $\int_0^{1} [f(1)-f(s)] dBs$
how could I prove this and would there be a general formula for $\int_0^{x_1} f'(s) B_s ds$ when $x_1 \neq 1$?
The second relation is just a restatement of the first one; indeed, one has : $$ \int_0^t f'(s) \,\mathrm{d}B_s = f(t)B_t - \int_0^t f(s) \,\mathrm{d}B_s = f(t)\int_0^t\mathrm{d}B_s - \int_0^t f(s) \,\mathrm{d}B_s = \int_0^t f(t)-f(s) \,\mathrm{d}B_s $$ with $t=1$ in your example.
It is to be noted that, in practice, we much prefer to stay with the expression $\int_0^tf(s)\mathrm{d}B_s$ nonetheless, because we know straightforwardly that it is distributed as $\mathcal{N}\left(0,\int_0^tf(s)^2\mathrm{d}s\right)$.