If int frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}} d | vedclass

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Question

(D) Let $I = \int \frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}} dx$. Multiply numerator and denominator by $e^x$: $I = \int \frac{4e^{2x} + 6}{9e^{2x} - 4} dx

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let $I = \int \frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}} dx$. Multiply numerator and denominator by $e^x$: $I = \int \frac{4e^{2x} + 6}{9e^{2x} - 4} dx$. We express the numerator as $4e^{2x} + 6 = m(18e^{2x}) + n(9e^{2x} - 4)$. Comparing coefficients: $18m + 9n = 4$ and $-4n = 6 \implies n = -\frac{3}{2}$. $18m + 9(-\frac{3}{2}) = 4 \implies 18m = 4 + \frac{27}{2} = \frac{35}{2} \implies m = \frac{35}{36}$. Thus,$I = \int \frac{\frac{35}{36}(18e^{2x}) - \frac{3}{2}(9e^{2x} - 4)}{9e^{2x} - 4} dx = \frac{35}{36} \int \frac{18e^{2x}}{9e^{2x} - 4} dx - \frac{3}{2} \int dx$. $I = \frac{35}{36} \log |9e^{2x} - 4| - \frac{3}{2}x + C$. Comparing with $Ax + B \log |9e^{2x} - 4| + C$,we get $A = -\frac{3}{2}$ and $B = \frac{35}{36}$. Since none of the options match these values,the correct answer is $(d)$.

If $\int \frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}} dx = Ax + B \log |9e^{2x} - 4| + C$,then $A, B$ and $C$ are

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