If int frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d | vedclass

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Question

(A) Let $I = \int \frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d x$. Multiply numerator and denominator by $e^x$: $I = \int \frac{2 e^{2x}+3}{3 e^{2x}+4} d x

Vedclass · Accepted Answer

$\frac{3}{4}, \frac{-1}{24}$

Vedclass · Answer

(A) Let $I = \int \frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d x$. Multiply numerator and denominator by $e^x$: $I = \int \frac{2 e^{2x}+3}{3 e^{2x}+4} d x$. We want to express the numerator in terms of the denominator and its derivative. Let $2 e^{2x} + 3 = A_1(3 e^{2x} + 4) + B_1 \frac{d}{dx}(3 e^{2x} + 4)$. $2 e^{2x} + 3 = 3 A_1 e^{2x} + 4 A_1 + 6 B_1 e^{2x}$. Comparing coefficients: $3 A_1 + 6 B_1 = 2$ and $4 A_1 = 3 \implies A_1 = \frac{3}{4}$. Substituting $A_1$: $3(\frac{3}{4}) + 6 B_1 = 2 \implies \frac{9}{4} + 6 B_1 = 2 \implies 6 B_1 = 2 - \frac{9}{4} = -\frac{1}{4} \implies B_1 = -\frac{1}{24}$. Thus,$I = \int \left( \frac{3}{4} - \frac{1}{24} \frac{6 e^{2x}}{3 e^{2x} + 4} \right) d x$. $I = \frac{3}{4} x - \frac{1}{24} \log(3 e^{2x} + 4) + C$. Comparing with $Ax + B \log(3 e^{2x} + 4) + C$,we get $A = \frac{3}{4}$ and $B = -\frac{1}{24}$.

If $\int \frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d x=A x+B \log \left(3 e^{2 x}+4\right)+C$,then values of $A$ and $B$ are respectively (where $C$ is a constant of integration.)

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