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$\int \frac{4 e^{x}+6 e^{-x}}{9 e^{x}-4 e^{-x}} d x=A x+B \log \left|9 e^{2 x}-4\right|+c$, then
(Where $c$ is constant of integration)
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(Where $c$ is constant of integration)
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Verified Answer
The correct answer is:
$A=\frac{-3}{2}, B=\frac{35}{36}$
$I=\int \frac{4 e^{2 x}+6}{9 e^{2 x}-4} d x$
$4 e^{2 x}+6=A\left(18 e^{2 x}\right)+B\left(9 e^{2 x}-4\right)...(1)$
$4 e^{2 x}+6=(18 A+9 B) e^{2 x}-4 B$
$\therefore-4 \mathrm{~B}=6 \Rightarrow \mathrm{B}=\frac{-3}{2}$ and $18 \mathrm{~A}+9 \mathrm{~B}=4$
$\therefore \quad 18 \mathrm{~A}-\frac{9 \times 3}{2}=4 \Rightarrow 18 \mathrm{~A}=\frac{35}{2} \Rightarrow \mathrm{A}=\frac{35}{36}$
$\therefore 4 e^{2 x}+6=\frac{35}{36}\left(18 e^{2 x}\right)-\frac{3}{2}\left(9 e^{2 x}-4\right)$
$I=\int\left[\frac{\frac{35}{36}\left(18 e^{2 x}\right)}{9 e^{2 x}-4}-\frac{\frac{3}{2}\left(9 e^{x}-4\right)}{9 e^{x}-4}\right] d x$
$I=\frac{35}{36} \log \left|9 e^{2 x}-4\right|-\frac{3}{2} x+c=A x+B \log \left|9 e^{2 x}-4\right|+c$
$\Rightarrow \mathrm{A}=-\frac{3}{2}$ and $\mathrm{B}=\frac{35}{36}$
$4 e^{2 x}+6=A\left(18 e^{2 x}\right)+B\left(9 e^{2 x}-4\right)...(1)$
$4 e^{2 x}+6=(18 A+9 B) e^{2 x}-4 B$
$\therefore-4 \mathrm{~B}=6 \Rightarrow \mathrm{B}=\frac{-3}{2}$ and $18 \mathrm{~A}+9 \mathrm{~B}=4$
$\therefore \quad 18 \mathrm{~A}-\frac{9 \times 3}{2}=4 \Rightarrow 18 \mathrm{~A}=\frac{35}{2} \Rightarrow \mathrm{A}=\frac{35}{36}$
$\therefore 4 e^{2 x}+6=\frac{35}{36}\left(18 e^{2 x}\right)-\frac{3}{2}\left(9 e^{2 x}-4\right)$
$I=\int\left[\frac{\frac{35}{36}\left(18 e^{2 x}\right)}{9 e^{2 x}-4}-\frac{\frac{3}{2}\left(9 e^{x}-4\right)}{9 e^{x}-4}\right] d x$
$I=\frac{35}{36} \log \left|9 e^{2 x}-4\right|-\frac{3}{2} x+c=A x+B \log \left|9 e^{2 x}-4\right|+c$
$\Rightarrow \mathrm{A}=-\frac{3}{2}$ and $\mathrm{B}=\frac{35}{36}$
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