Given linear mass density

$\rho(x)=a+b\left(\frac{x}{L}\right)^2, \qquad 0\le x\le L$

For a rod along the $x$-axis, the centre of mass is

$x_{\text{cm}}=\frac{\int_0^L x\,\rho(x)\,dx}{\int_0^L \rho(x)\,dx}$

Compute total mass

$M=\int_0^L \rho(x)\,dx=\int_0^L \left(a+b\frac{x^2}{L^2}\right)dx$

$M=a\int_0^L dx+\frac{b}{L^2}\int_0^L x^2 dx$

$M=aL+\frac{b}{L^2}\cdot \frac{L^3}{3}$

$M=L\left(a+\frac{b}{3}\right)$

Compute numerator for centre of mass

$\int_0^L x\rho(x)dx=\int_0^L x\left(a+b\frac{x^2}{L^2}\right)dx$

$=a\int_0^L x\,dx+\frac{b}{L^2}\int_0^L x^3 dx$

$=a\cdot \frac{L^2}{2}+\frac{b}{L^2}\cdot \frac{L^4}{4}$

$=L^2\left(\frac{a}{2}+\frac{b}{4}\right)$

Find $x_{\text{cm}}$

$x_{\text{cm}}=\frac{L^2\left(\frac{a}{2}+\frac{b}{4}\right)}{L\left(a+\frac{b}{3}\right)}$

$=L\cdot \frac{\frac{2a+b}{4}}{\frac{3a+b}{3}}$

$=L\cdot \frac{2a+b}{4}\cdot \frac{3}{3a+b}$

$x_{\text{cm}}=\frac{3}{4}\left(\frac{2a+b}{3a+b}\right)L$

Match with options

This is exactly Option C:

$\boxed{\frac{3}{4}\left(\frac{2a+b}{3a+b}\right)L}$

Comparison with stored answer

Stored correct answer: C
Derived answer: C
Hence, they agree.