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

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Aicharya · Accepted Answer

$I=\int \frac{4 e^{2 x}+6}{9 e^{2 x}-4} d x$ $4 e^{2 x}+6=A\left(18 e^{2 x}ight)+B\left(9 e^{2 x}-4ight)...(1)$ $4 e^{2 x}+6=(18 A+9 B) e^{2 x}-4 B$ $	herefore-4 \mathrm{~B}=6 \Rightarrow \mathrm{B}=\frac{-3}{2}$ and $18 \mathrm{~A}+9 \mathrm{~B}=4$ $	herefore \quad 18 \mathrm{~A}-\frac{9 	imes 3}{2}=4 \Rightarrow 18 \mathrm{~A}=\frac{35}{2} \Rightarrow \mathrm{A}=\frac{35}{36}$ $	herefore 4 e^{2 x}+6=\frac{35}{36}\left(18 e^{2 x}ight)-\frac{3}{2}\left(9 e^{2 x}-4ight)$ $I=\int\left[\frac{\frac{35}{36}\left(18 e^{2 x}ight)}{9 e^{2 x}-4}-\frac{\frac{3}{2}\left(9 e^{x}-4ight)}{9 e^{x}-4}ight] d x$ $I=\frac{35}{36} \log \left|9 e^{2 x}-4ight|-\frac{3}{2} x+c=A x+B \log \left|9 e^{2 x}-4ight|+c$ $\Rightarrow \mathrm{A}=-\frac{3}{2}$ and $\mathrm{B}=\frac{35}{36}$

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

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