A thin uniformly charged rod of length $L$ lies along the $x$-axis from $x=0$ to $x=L$. Point $P$ is located on the $x$-axis at $x=L+a$ (with $a>0$). If the rod has linear charge density $\lambda$, select the integral that gives the electric potential at $P$.

$V\left(P\right)=k{\int }_0^L\frac{\lambda dx}{L+a+x}$

$V\left(P\right)=k{\int }_0^L\frac{\lambda dx}{(L+a-x{)}^2}$

$V\left(P\right)=k{\int }_0^L\frac{\lambda dx}{L+a-x}$

$V\left(P\right)=k{\int }_0^L\lambda (L+a-x)dx$