If int frac{3e^x - 5e^{-x}}{4e^x + 5e^{-x}} d | vedclass

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Question

(D) Let $I = \int \frac{3e^x - 5e^{-x}}{4e^x + 5e^{-x}} dx$. We express the numerator as $A(4e^x - 5e^{-x}) + B(4e^x + 5e^{-x})$,where $4e^x - 5e^{-x}

Vedclass · Accepted Answer

$p = -1/8, q = 7/8$

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(D) Let $I = \int \frac{3e^x - 5e^{-x}}{4e^x + 5e^{-x}} dx$. We express the numerator as $A(4e^x - 5e^{-x}) + B(4e^x + 5e^{-x})$,where $4e^x - 5e^{-x}$ is the derivative of the denominator $4e^x + 5e^{-x}$. Equating coefficients of $e^x$ and $e^{-x}$: $4A + 4B = 3$ $-5A + 5B = -5$ From the second equation,$-A + B = -1$,so $B = A - 1$. Substituting into the first: $4A + 4(A - 1) = 3 \implies 8A - 4 = 3 \implies 8A = 7 \implies A = 7/8$. Then $B = 7/8 - 1 = -1/8$. So,$I = \int \frac{7/8(4e^x - 5e^{-x}) - 1/8(4e^x + 5e^{-x})}{4e^x + 5e^{-x}} dx$. $I = 7/8 \int \frac{4e^x - 5e^{-x}}{4e^x + 5e^{-x}} dx - 1/8 \int 1 dx$. $I = 7/8 \log |4e^x + 5e^{-x}| - 1/8 x + C$. Comparing with $px + q \log |4e^x + 5e^{-x}| + C$,we get $p = -1/8$ and $q = 7/8$.

If $\int \frac{3e^x - 5e^{-x}}{4e^x + 5e^{-x}} dx = px + q \cdot \log |4e^x + 5e^{-x}| + C$,then

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