Hello. Have a nice day. I would like to get the solution of this integral step by step please. I would like each step to be very detailed please, because I am not very good with this type of integrals. The exercise is from Mcquarrie's quantum chemistry book, problem $8-18,$ and this is located at Chegg.

$E(Z)=d{r}_{1}d{r}_2{}^{**}$hat$(H)$

$=d{r}_{1}d{r}_2{}^{**}[(-\frac{e^2{Z}^2}{4{}_0})+\frac{(Z-2)e^2}{4{}_0{r}_1}+\frac{(Z-2)e^2}{4{}_0{r}_2}+\frac{e^2}{4{}_0{r}_{12}}]$

$=\frac{Z^6}{a_0^6{}^2}d{r}_{1}d{r}_2[(-\frac{e^2{Z}^2}{4{}_0})+\frac{(Z-2)e^2}{4{}_0{r}_1}+\frac{(Z-2)e^2}{4{}_0{r}_2}+\frac{e^2}{4{}_0{r}_{12}}]e^{\frac{-2Z(r_1+r_2)}{a_0}}$

Now this can be broken into four integrals

$\frac{Z^6}{a_0^6{}^2}d{r}_{1}d{r}_2[(-\frac{e^2{Z}^2}{4{}_0})+\frac{(Z-2)e^2}{4{}_0{r}_1}+\frac{(Z-2)e^2}{4{}_0{r}_2}+\frac{e^2}{4{}_0{r}_{12}}]e^{\frac{-2Z(r_1+r_2)}{a_0}}=-Z^2(\frac{m_e{e}^4}{16{}^2{}_0^2{}^2})$

$\frac{Z^6}{a_0^6{}^2}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}d{r}_1[\frac{(Z-2)e^2}{4{}_0{r}_1}]e^{\frac{-2Z_1}{a_0}}=Z(Z-2)(\frac{m_e{e}^4}{16{}^2{}_0^2{}^2})$

$\frac{Z^6}{a_0^6{}^2}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}d{r}_2[\frac{(Z-2)e^2}{4{}_0{r}_2}]e^{\frac{-2Z_2}{a_0}}=Z(Z-2)(\frac{m_e{e}^4}{16{}^2{}_0^2{}^2})$ and

$\frac{Z^6}{a_0^6{}^2}d{r}_{1}d{r}_2[\frac{e^2}{4{}_0{r}_{12}}]e^{\frac{-2Z({}_1+r_2)}{a_0}}=\frac{5Z}{8}(\frac{m_e{e}^4}{16{}^2{}_0^2{}^2})<mi\; id="EditorElement\_3256850"></mi>$

Therefore, in units of

$\frac{m_e{e}^4}{16{}^2{}_0^2{}^2}$