By changing variables as in the text, express the diagram in equation $8.18$ in terms of the same integral as in the equation$8.31$. Do the same for the last two diagrams in the first line of the equation$8.20$. Which diagrams cannot be written in terms of this basic integral?

We need to find that the diagrams cannot be written in terms of this basic integral.

For the diagram in the equation $8.18$,we have three dots, so first we multiply by $N$then $N-1$then $N-2$, and for the two lines connected them we multiply by $f_{12}{f}_{23}$and for each dot we have $\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$V$}\right.\right)\int d^3{r}_i$, the symmetry factor is $2$, since we can number the first dot with $1$and third dot with $3$and vice versa so we have two ways, so write the second diagram as :

$\left(3\right)=\frac{1}{2}\frac{N\left(N-1\right)\left(N-2\right)}{V^3}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}$

let $N\approx N-1\approx N-2$

$\left(3\right)=\frac{1}{2}\frac{N^3}{V^3}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}$

the center dot be $1$,so we can write the integer as follow

$\left(3\right)=\frac{1}{2}\frac{N^3}{V^3}\int d^3{r}_1{\left(\int d^{3}rf\left(r\right)\right)}^2$

Here,

$f\left(r\right)=e^{-\beta u\left(r\right)}-1$

the result of the integral over $r$is the intensive quantity that is independent of $r_1$and $V$, because $f\left(r\right)$goes to zero when $r$is only a few times larger than the size of the molecule, so whatever the value of this integral, the remaining integral over $r_1$results the factor of $V$,

$\left(3\right)=\frac{1}{2}\frac{N^3}{V^3}{\left(\int d^{3}rf\left(r\right)\right)}^2$

We can not express the last two diagrams in equation $8.20$,but we can still determine the size, let $v$be the size of the molecule, so the first diagram will have a size of :

$\frac{N^4{v}^3}{V^3}$

and also the second one will have the same size.