Answer: \(\tanh^{-1}\!\left(\frac12 e^x\right)=\frac12\ln 3+\frac23x+\frac59x^2+\cdots\)

So \(a=3\), \(b=\frac23\), \(c=\frac59\).

Let \(f(x)=\tanh^{-1}\!\left(\frac12 e^x\right)\).

The Maclaurin expansion up to the \(x^2\) term is

\(f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\cdots\)

First, using \(\dfrac{d}{dx}(\tanh^{-1}u)=\dfrac{u'}{1-u^2}\) with \(u=\dfrac12 e^x\),

\(f'(x)=\frac{\frac12 e^x}{1-\left(\frac12 e^x\right)^2}=\frac{\frac12 e^x}{1-\frac14 e^{2x}}\).

So

\(f'(0)=\frac{\frac12}{1-\frac14}=\frac{2}{3}.\)

Differentiate again:

\(f''(x)=\frac{\left(1-\frac14 e^{2x}\right)\left(\frac12 e^x\right)-\frac12 e^x\left(-\frac12 e^{2x}\right)}{\left(1-\frac14 e^{2x}\right)^2}.\)

Hence

\(f''(0)=\frac{\left(1-\frac14\right)\left(\frac12\right)-\frac12\left(-\frac12\right)}{\left(1-\frac14\right)^2}=\frac{10}{9}.\)

Now

\(f(0)=\tanh^{-1}\!\left(\frac12\right).\)

Using \(\tanh^{-1}t=\frac12\ln\!\left(\frac{1+t}{1-t}\right)\),

\(f(0)=\frac12\ln\!\left(\frac{1+\frac12}{1-\frac12}\right)=\frac12\ln\!\left(\frac{\frac32}{\frac12}\right)=\frac12\ln 3.\)

Therefore

\(f(x)=\frac12\ln 3+\frac23x+\frac12\cdot\frac{10}{9}x^2+\cdots\)

\(\phantom{f(x)}=\frac12\ln 3+\frac23x+\frac59x^2+\cdots\)

So in the form \(\frac12\ln a+bx+cx^2\),

\(a=3\), \(b=\frac23\), \(c=\frac59\).