Divide a rational function and integrate an inverse-tangent product

Divide first: $(8x^2+1)/(4x^2-1)=2+3/(4x^2-1)$. Either use the standard difference-of-squares integral or the printed partial fractions $\dfrac1{4x^2-1}=\dfrac1{2(2x-1)}-\dfrac1{2(2x+1)}$. This gives $2x+\frac34\ln\left|\dfrac{2x-1}{2x+1}\right|+C$, equivalently $2x+\frac34\ln|2x-1|-\frac34\ln|2x+1|+C$.

Choose $u=\tan^{-1}(e^{-3x})$, $dv=e^{3x}dx$. Then $du=-3e^{-3x}/(1+e^{-6x})\,dx$ and $v=e^{3x}/3$. Integration by parts gives $\frac13e^{3x}\tan^{-1}(e^{-3x})+\int\dfrac1{1+e^{-6x}}dx$. As the source callout explains, multiply numerator and denominator by $e^{6x}$ to get $\int e^{6x}/(e^{6x}+1)dx=\frac16\ln(e^{6x}+1)$. Therefore the full antiderivative is $\frac13e^{3x}\tan^{-1}(e^{-3x})+\frac16\ln(e^{6x}+1)+C$.