as requested by the OP in the comment section:

$$\int_0^T e^{-x}I_n(x)\frac{1}{x}\,dx=\frac{1}{n T^{n-1}}e^{-T}\left[e^T T^{n-1}+a_n(T)I_0(T)+b_n(T)I_1(T)\right]$$

the functions $a_n$ and $b_n$ are polynomials of degree $n-1$, I do not have a closed form expression; the first few are:

$$a_1(T)=-1,\;\;a_2(T)=-2T,\;\;a_3(T)=-3 T^2+4 T,$$ $$a_4(T)=-4 T^3+8 T^2-24 T,\;\;a_5(T)=-5 T^4+20 T^3-48 T^2+192 T$$ $$b_1(T)=-1,\;\;b_2(T)=-2T+2,\;\;b_3(T)=-3T^2+4T-8,$$ $$b_4(T)=-4 T^3+12 T^2-16 T+48,\;\;b_5(T)=-5 T^4+20 T^3-88 T^2+96 T-384$$

as requested by the OP in the comment section:

$$\int_0^T e^{-x}I_n(x)\frac{1}{x}\,dx=\frac{1}{n}+\frac{1}{n T^{n-1}}e^{-T}\left[a_n(T)I_0(T)+b_n(T)I_1(T)\right]$$

the functions $a_n$ and $b_n$ are polynomials of degree $n-1$, I do not have a closed form expression; the first few are:

$$a_1(T)=-1,\;\;a_2(T)=-2T,\;\;a_3(T)=-3 T^2+4 T,$$ $$a_4(T)=-4 T^3+8 T^2-24 T,\;\;a_5(T)=-5 T^4+20 T^3-48 T^2+192 T$$ $$b_1(T)=-1,\;\;b_2(T)=-2T+2,\;\;b_3(T)=-3T^2+4T-8,$$ $$b_4(T)=-4 T^3+12 T^2-16 T+48,\;\;b_5(T)=-5 T^4+20 T^3-88 T^2+96 T-384$$