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

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Question

(C) Multiply numerator and denominator by $e^x$: $I = \int \frac{4 e^{2x} + 6}{9 e^{2x} - 4} dx$ Let $4 e^{2x} + 6 = A(18 e^{2x}) + B(9 e^{2x} - 4)$ C

Vedclass · Accepted Answer

$A=\frac{-3}{2}, B=\frac{35}{36}$

Vedclass · Answer

(C) Multiply numerator and denominator by $e^x$: $I = \int \frac{4 e^{2x} + 6}{9 e^{2x} - 4} dx$ Let $4 e^{2x} + 6 = A(18 e^{2x}) + B(9 e^{2x} - 4)$ Comparing coefficients of $e^{2x}$ and constants: $18A + 9B = 4$ and $-4B = 6$ From $-4B = 6$,we get $B = -\frac{3}{2}$ Substituting $B$ in $18A + 9B = 4$: $18A + 9(-\frac{3}{2}) = 4 \Rightarrow 18A - \frac{27}{2} = 4 \Rightarrow 18A = \frac{35}{2} \Rightarrow A = \frac{35}{36}$ Now,$I = \int \left[ \frac{\frac{35}{36}(18 e^{2x})}{9 e^{2x} - 4} - \frac{\frac{3}{2}(9 e^{2x} - 4)}{9 e^{2x} - 4} \right] dx$ $I = \frac{35}{36} \log |9 e^{2x} - 4| - \frac{3}{2} x + c$ Comparing with $Ax + B \log |9 e^{2x} - 4| + c$,we get $A = -\frac{3}{2}$ and $B = \frac{35}{36}$.

$\int \frac{4 e^{x}+6 e^{-x}}{9 e^{x}-4 e^{-x}} d x=A x+B \log \left|9 e^{2 x}-4\right|+c$,then (Where $c$ is constant of integration)

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