If the linear density of the rod of length L varies as $\lambda =A+Bx$ , then its centre of mass is given by

As the rod is along $x$ -axis, for all points on it $y$  and $z$  will be zero,

so,  $Y_{\text{CM}}$  = 0 and  $Z_{\text{CM}}$   = 0

i.e., the centre of mass will lie on the rod.  Now, consider an element of rod of length $dx$  at a distance $x$  from the origin, then

$dm$  = $\lambda$   $dx$  = $(A+Bx)$ $dx$

So,   ${\text{X}}_{\text{CM}}$  = $\frac{{\displaystyle {\int }_0^{L}xdm}}{{\displaystyle {\int }_0^{L}dm}}=\frac{{\displaystyle {\int }_0^{L}x(A+Bx)dx}}{{\displaystyle {\int }_0^L(A+Bx)dx}}$

$\frac{\frac{A{L}^2}{2}+\frac{B{L}^3}{3}}{AL\text{}\text{}+\frac{B{L}^2}{2}}=\frac{L(3A+2BL)}{3(2A+BL)}$