If lim _ t → 0 (∫_0^1(3 x+5)^ t d x )^ 1 t= α 5 e ( 8 5 )^ 2 3, then α is equal to

$\displaystyle \begin{aligned} \frac{\alpha}{5 e} & =\exp \left(\lim _{t \rightarrow 0} \frac{1}{t}\left(\int_0^1(3 x+5)^t d x-1\right)\right) \\ & =\exp \left(\lim _{t \rightarrow 0} \frac{1}{t}\left(\frac{(3 x+5)^{t+1}}{3(t+1)}\right)_0^1-1\right) \\ & =\exp \left(\lim _{t \rightarrow 0} \frac{1}{t}\left(\frac{8^{t+1}-5^{t+1}}{3(t+1)}-1\right)\right) \\ & =\exp \left(\lim _{t \rightarrow 0} \frac{1}{t}\left(\frac{8^{t+1}-5^{t+1}-3 t-3}{3(t+1)}\right)\right) \\ & =\exp \left(\lim _{t \rightarrow 0}\left(\frac{8^{t+1} \cdot \ln 8-5^{t+1} \ln 5-3}{3(t+1)}\right)\right) \\ & =\exp \left(\frac{\ln 8^8-\ln 5^5-3}{5}\right)\end{aligned}$5eα​​=exp(t→0lim​t1​(∫01​(3x+5)tdx−1))=exp(t→0lim​t1​(3(t+1)(3x+5)t+1​)01​−1)=exp(t→0lim​t1​(3(t+1)8t+1−5t+1​−1))=exp(t→0lim​t1​(3(t+1)8t+1−5t+1−3t−3​))=exp(t→0lim​(3(t+1)8t+1⋅ln8−5t+1ln5−3​))=exp(5ln88−ln55−3​)​ $\begin{aligned} & =\left(\frac{8}{5}\right)^{2 / 3} \frac{\alpha}{5 e}=\exp \left(\frac{\ln \left(\frac{8^8}{5^5}\right)}{5}-1\right) \\ & \Rightarrow\left(\frac{8}{5}\right)^{2 / 3} \frac{\alpha}{5}=\left(\frac{8^8}{5^5}\right)^{1 / 3}=\left(\frac{8^6 \cdot 8^2}{5^3 \cdot 5^2}\right)^{1 / 3}=\frac{64}{5}\left(\frac{8}{5}\right)^{2 / 3} \\ & \Rightarrow \alpha=64\end{aligned}$​=(58​)2/35eα​=exp​5ln(5588​)​−1​⇒(58​)2/35α​=(5588​)1/3=(53⋅5286⋅82​)1/3=564​(58​)2/3⇒α=64​