1. **State the problem:** We are given two functions: $$f(x) = \frac{e^{2x} + 2e^{x} + 2}{e^{x} + 1}$$ and $$g(x) = \ln|x|$$ We need to find the inverse of the composition $(f \circ g)^{-1}(x)$, which means find the inverse function of $f(g(x))$. 2. **Write the composition:** $$ (f \circ g)(x) = f(g(x)) = f(\ln|x|) = \frac{e^{2\ln|x|} + 2e^{\ln|x|} + 2}{e^{\ln|x|} + 1} $$ Recall that $e^{\ln|x|} = |x|$, so: $$ (f \circ g)(x) = \frac{|x|^{2} + 2|x| + 2}{|x| + 1} $$ 3. **Simplify the numerator:** $$ |x|^{2} + 2|x| + 2 = (|x| + 1)^{2} + 1 $$ So, $$ (f \circ g)(x) = \frac{(|x| + 1)^{2} + 1}{|x| + 1} $$ 4. **Divide numerator by denominator:** $$ (f \circ g)(x) = \frac{\cancel{(|x| + 1)}(|x| + 1) + 1}{\cancel{|x| + 1}} = |x| + 1 + \frac{1}{|x| + 1} $$ 5. **Set $y = (f \circ g)(x)$:** $$ y = |x| + 1 + \frac{1}{|x| + 1} $$ Let $t = |x| + 1 > 0$, then: $$ y = t + \frac{1}{t} $$ 6. **Solve for $t$:** Multiply both sides by $t$: $$ yt = t^{2} + 1 $$ Rearranged: $$ t^{2} - yt + 1 = 0 $$ 7. **Use quadratic formula:** $$ t = \frac{y \pm \sqrt{y^{2} - 4}}{2} $$ Since $t = |x| + 1 > 0$, we take the positive root: $$ t = \frac{y + \sqrt{y^{2} - 4}}{2} $$ 8. **Find $|x|$:** $$ |x| = t - 1 = \frac{y + \sqrt{y^{2} - 4}}{2} - 1 $$ 9. **Find $x$:** Since $g(x) = \ln|x|$, the domain of $g$ is $x \neq 0$, and $f \circ g$ depends on $|x|$, the inverse will have two branches: $$ x = \pm \left( \frac{y + \sqrt{y^{2} - 4}}{2} - 1 \right) $$ 10. **Express the inverse function:** $$ (f \circ g)^{-1}(x) = \pm \left( \frac{x + \sqrt{x^{2} - 4}}{2} - 1 \right) $$ **Final answer:** $$ (f \circ g)^{-1}(x) = \pm \left( \frac{x + \sqrt{x^{2} - 4}}{2} - 1 \right) $$ This gives the inverse of the composition function for $x \geq 2$ (since $y^{2} - 4 \geq 0$ for real values).