Expand a logarithm after changing its argument

The Student prints the first factorisation; the Teacher completes the solution. Write $2+x^2=2(1+x^2/2)$, hence $\ln(2+x^2)=\ln2+\ln(1+x^2/2)$. The source explicitly uses $\ln(AB)=\ln A+\ln B$; the two factors are positive for real x.

Use $u=x^2/2$ in $\ln(1+u)=u-u^2/2+u^3/3-\cdots$. To retain terms through $x^6$, keep through $u^3$: $\ln(2+x^2)=\ln2+(x^2/2)-(x^2/2)^2/2+(x^2/2)^3/3-\cdots$.

The Teacher obtains $\ln(2+x^2)=\ln2+x^2/2-x^4/8+x^6/24-\cdots$. The first omitted nonzero power is $x^8$. The polynomial through $x^6$ is an approximation near zero; the infinite series interval is determined separately in Example 2.