int_{log 4}^{log 5} frac{e^{2 x}+e^x}{e^{2 x} | vedclass

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(D) Let $I = \int_{\log 4}^{\log 5} \frac{e^{2x} + e^x}{e^{2x} - 5e^x + 6} dx$. Substitute $e^x = t$,then $e^x dx = dt$. When $x = \log 4$,$t = 4$. Wh

Vedclass · Accepted Answer

$\log \left(\frac{128}{27}\right)$

Vedclass · Answer

(D) Let $I = \int_{\log 4}^{\log 5} \frac{e^{2x} + e^x}{e^{2x} - 5e^x + 6} dx$. Substitute $e^x = t$,then $e^x dx = dt$. When $x = \log 4$,$t = 4$. When $x = \log 5$,$t = 5$. $I = \int_4^5 \frac{t+1}{(t-3)(t-2)} dt$. Using partial fractions: $\frac{t+1}{(t-3)(t-2)} = \frac{A}{t-3} + \frac{B}{t-2}$. $t+1 = A(t-2) + B(t-3)$. For $t=3$,$4 = A(1) \Rightarrow A = 4$. For $t=2$,$3 = B(-1) \Rightarrow B = -3$. $I = \int_4^5 \left( \frac{4}{t-3} - \frac{3}{t-2} \right) dt$. $I = [4 \log|t-3| - 3 \log|t-2|]_4^5$. $I = (4 \log 2 - 3 \log 3) - (4 \log 1 - 3 \log 2)$. $I = 4 \log 2 - 3 \log 3 + 3 \log 2 = 7 \log 2 - 3 \log 3$. $I = \log(2^7) - \log(3^3) = \log\left(\frac{128}{27}\right)$.

$\int_{\log 4}^{\log 5} \frac{e^{2 x}+e^x}{e^{2 x}-5 e^x+6} d x=$

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