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

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(D) दिया गया है $\alpha = \int_0^1 e^{(9x + 3 	an^{-1} x)} \left(\frac{12 + 9x^2}{1 + x^2}ight) dx$. माना $t = 9x + 3 	an^{-1} x$. तब,$dt = \left(

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$9$

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(D) दिया गया है $\alpha = \int_0^1 e^{(9x + 3 	an^{-1} x)} \left(\frac{12 + 9x^2}{1 + x^2}ight) dx$. माना $t = 9x + 3 	an^{-1} x$. तब,$dt = \left(9 + \frac{3}{1 + x^2}ight) dx = \left(\frac{9(1 + x^2) + 3}{1 + x^2}ight) dx = \left(\frac{12 + 9x^2}{1 + x^2}ight) dx$. जब $x = 0$,तब $t = 9(0) + 3 	an^{-1}(0) = 0$. जब $x = 1$,तब $t = 9(1) + 3 	an^{-1}(1) = 9 + 3(\frac{\pi}{4}) = 9 + \frac{3\pi}{4}$. अतः,$\alpha = \int_0^{9 + \frac{3\pi}{4}} e^t dt = [e^t]_0^{9 + \frac{3\pi}{4}} = e^{9 + \frac{3\pi}{4}} - e^0 = e^{9 + \frac{3\pi}{4}} - 1$. इसलिए,$1 + \alpha = e^{9 + \frac{3\pi}{4}}$. प्राकृतिक लघुगणक लेने पर,$\log_e |1 + \alpha| = 9 + \frac{3\pi}{4}$. अंत में,$\log_e |1 + \alpha| - \frac{3\pi}{4} = 9 + \frac{3\pi}{4} - \frac{3\pi}{4} = 9$.

यदि $\alpha = \int_0^1 \left(e^{9x + 3 \tan^{-1} x}\right) \left(\frac{12 + 9x^2}{1 + x^2}\right) dx$,जहाँ $\tan^{-1} x$ केवल मुख्य मान लेता है,तो $\left(\log_e |1 + \alpha| - \frac{3\pi}{4}\right)$ का मान है

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