Answer: \(\ln(x+2)+\ln(x^2+5)=\ln 10+\frac{1}{2}x+\frac{3}{40}x^2\)

Use the standard Maclaurin expansion \(\ln(1+u)=u-\frac{u^2}{2}+\cdots\).

First,

\(\ln(x+2)=\ln\!\left(2\left(1+\frac{x}{2}\right)\right)=\ln 2+\ln\!\left(1+\frac{x}{2}\right)\).

So

\(\ln(x+2)=\ln 2+\frac{x}{2}-\frac{1}{2}\left(\frac{x}{2}\right)^2+\cdots=\ln 2+\frac{1}{2}x-\frac{1}{8}x^2+\cdots\).

Next,

\(\ln(x^2+5)=\ln\!\left(5\left(1+\frac{x^2}{5}\right)\right)=\ln 5+\ln\!\left(1+\frac{x^2}{5}\right)\).

Hence

\(\ln(x^2+5)=\ln 5+\frac{x^2}{5}+\cdots\),

since the next term is of order \(x^4\).

Now add the two series:

\(\ln(x+2)+\ln(x^2+5)=\left(\ln 2+\frac{1}{2}x-\frac{1}{8}x^2+\cdots\right)+\left(\ln 5+\frac{1}{5}x^2+\cdots\right)\).

Combining like terms gives

\(\ln 2+\ln 5=\ln 10\), and \(-\frac{1}{8}+\frac{1}{5}=\frac{3}{40}\).

Therefore, up to and including the term in \(x^2\),

\(\ln(x+2)+\ln(x^2+5)=\ln 10+\frac{1}{2}x+\frac{3}{40}x^2\).