If $\alpha = \int_{0}^{1} \left(e^{9x+3\tan^{-1}x}\right) \left(\frac{12+9x^{2}}{1+x^{2}}\right) dx$ where $\tan^{-1}x$ takes only principal values, then the value of $\left(\log_e |1+\alpha| - \frac{3\pi}{4}\right)$ is

If α = ∫₀¹ (e^(9x+3tan⁻¹x)) ((12+9x²)/(1+x²)) dx where tan⁻¹x takes only principal values, then the value of (logₑ |1+α| - 3π/4) is $\alpha = \int_{0}^{1} e^{9x+3\tan^{-1}x}\left(\frac{12+9x^{2}}{1+x^{2}}\right)dx$ $\text{Find } \alpha \ \text{then } \log_e|1+\alpha| - \frac{3\pi}{4}$ $1.\ g(x) = 9x+3\tan^{-1}x \Rightarrow g'(x)$ $2.\ \alpha = \int e^{g}g'\,dx = \left[e^{g}\right]$ $3.\ \text{use limits } \Rightarrow \log_e|1+\alpha| - \frac{3\pi}{4}$ $\frac{d}{dx}\left(9x + 3\tan^{-1}x\right) = 9 + \frac{3}{1+x^2}$ $= \frac{9(1+x^2)+3}{1+x^2} = \frac{12+9x^2}{1+x^2}$ $\alpha = \int_0^1 e^{g(x)} g'(x)\,dx,\quad g(x)=9x+3\tan^{-1}x$ $= \left[e^{g(x)}\right]_0^1 = e^{g(1)} - e^{g(0)}$ $g(1) = 9 + 3\cdot\frac{\pi}{4} = 9 + \frac{3\pi}{4},\qquad g(0)=0$ $\alpha = e^{9+3\pi/4} - 1$ $1+\alpha = e^{9+3\pi/4} \Rightarrow \log_e|1+\alpha| = 9 + \frac{3\pi}{4}$ $\log_e|1+\alpha| - \frac{3\pi}{4} = 9$