Evaluate Logarithmic Expression Involving Definite Integral with Inverse Tangent

To evaluate the given logarithmic expression involving the definite integral, we begin by simplifying the integral expression for $\alpha$.

The given integral is: $\alpha = \int_{0}^{1} \left(e^{9x + 3\tan^{-1} x}\right) \left(\frac{12 + 9x^2}{1 + x^2}\right) dx$

We can rewrite the rational factor in the integrand as follows: $\frac{12 + 9x^2}{1 + x^2} = \frac{9(1 + x^2) + 3}{1 + x^2} = 9 + \frac{3}{1 + x^2}$

Substituting this back into the integral gives: $\alpha = \int_{0}^{1} e^{9x + 3\tan^{-1} x} \left( 9 + \frac{3}{1 + x^2} \right) dx$

Let $g(x) = 9x + 3\tan^{-1} x$. Differentiating $g(x)$ with respect to $x$, we obtain: $g'(x) = \frac{d}{dx}\left(9x + 3\tan^{-1} x\right) = 9 + \frac{3}{1 + x^2}$

Thus, the integrand is of the form $e^{g(x)} g'(x)$, whose antiderivative is simply $e^{g(x)}$: $\alpha = \int_{0}^{1} e^{g(x)} g'(x) \, dx = \left[ e^{9x + 3\tan^{-1} x} \right]_0^1$

Evaluating this at the upper limit $x = 1$ and lower limit $x = 0$: $\alpha = e^{9(1) + 3\tan^{-1}(1)} - e^{9(0) + 3\tan^{-1}(0)}$

Since $\tan^{-1}(1) = \frac{\pi}{4}$ and $\tan^{-1}(0) = 0$, we have: $\alpha = e^{9 + 3\left(\frac{\pi}{4}\right)} - e^0 = e^{9 + \frac{3\pi}{4}} - 1$

Rearranging the terms: $1 + \alpha = e^{9 + \frac{3\pi}{4}}$

Since $1 + \alpha > 0$, taking the natural logarithm on both sides yields: $\log_e |1 + \alpha| = \log_e \left( e^{9 + \frac{3\pi}{4}} \right) = 9 + \frac{3\pi}{4}$

Finally, substituting this into the expression we need to evaluate: $\left( \log_e |1 + \alpha| - \frac{3\pi}{4} \right) = \left( 9 + \frac{3\pi}{4} \right) - \frac{3\pi}{4} = 9$