If alpha = int_0^1 left(e^{9x + 3 tan^{-1} x} | vedclass

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(D) Given $\alpha = \int_0^1 e^{(9x + 3 	an^{-1} x)} \left(\frac{12 + 9x^2}{1 + x^2}ight) dx$. Let $t = 9x + 3 	an^{-1} x$. Then,$dt = \left(9 + \

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$9$

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(D) Given $\alpha = \int_0^1 e^{(9x + 3 	an^{-1} x)} \left(\frac{12 + 9x^2}{1 + x^2}ight) dx$. Let $t = 9x + 3 	an^{-1} x$. Then,$dt = \left(9 + \frac{3}{1 + x^2}ight) dx = \left(\frac{9(1 + x^2) + 3}{1 + x^2}ight) dx = \left(\frac{9 + 9x^2 + 3}{1 + x^2}ight) dx = \left(\frac{12 + 9x^2}{1 + x^2}ight) dx$. When $x = 0$,$t = 9(0) + 3 	an^{-1}(0) = 0$. When $x = 1$,$t = 9(1) + 3 	an^{-1}(1) = 9 + 3(\frac{\pi}{4}) = 9 + \frac{3\pi}{4}$. Thus,$\alpha = \int_0^{9 + \frac{3\pi}{4}} e^t dt = [e^t]_0^{9 + \frac{3\pi}{4}} = e^{9 + \frac{3\pi}{4}} - e^0 = e^{9 + \frac{3\pi}{4}} - 1$. Therefore,$1 + \alpha = e^{9 + \frac{3\pi}{4}}$. Taking the natural logarithm,$\log_e |1 + \alpha| = 9 + \frac{3\pi}{4}$. Finally,$\log_e |1 + \alpha| - \frac{3\pi}{4} = 9 + \frac{3\pi}{4} - \frac{3\pi}{4} = 9$.

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

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