"I am back" says Integration

Divide numerator and denominator by $e^{2x}$ to write it as: $\Large\displaystyle\int\dfrac{(e^x+e^{-x})\mathrm{d}x}{\underbrace{e^{2x}-1+e^{2x}}_{\Large\color{#D61F06}{(e^x-e^{-x})^2+1}}}$

Substitute $e^x-e^{-x}=t$ such that $(e^x+e^{-x})\mathrm{d}x=\mathrm{d}t$ .

$\Large \displaystyle\int\dfrac{\mathrm{d}t}{t^2+1}=\tan^{-1}t+C$

$\Large =\tan^{-1}(e^x-e^{-x})+C$

$\large \implies a=b=g=1, d=-1$

$\large \therefore 101(a^2+b+d^3+g^4)=\boxed{202}$