Mathematics

# If $\int {\frac{{4{e^x} + 6{e^{ - x}}}}{{9{e^x} - 4{e^{ - x}}}}dx} = Ax + B{\log _e}\left( {9{e^{2x}} - 4} \right) + C$ then $A=...... ,B=...... ,C=......$.

##### SOLUTION
$\displaystyle\int{\dfrac{4{e}^{x}+6{e}^{-x}}{9{e}^{x}-4{e}^{-x}}dx}$

$=\displaystyle\int{\dfrac{4{e}^{2x}+6}{9{e}^{2x}-4}dx}$

We can write $Numerator=A\left(Denominator\right)+B\dfrac{d}{dx}\left(Denominator\right)$

$4{e}^{2x}+6=A\left(9{e}^{2x}-4\right)+B\dfrac{d}{dx}\left(9{e}^{2x}-4\right)$

$4{e}^{2x}+6=A\left(9{e}^{2x}-4\right)+B\left(18{e}^{2x}\right)$

$4{e}^{2x}+6=\left(9A+18B\right){e}^{2x}+\left(-4A\right)$

Comparing the coefficients, we have

$-4A=6$ or $A=-\dfrac{6}{4}=-\dfrac{3}{2}$

$9A+18B=4\Rightarrow\,18B=4-9A=4+9\times\dfrac{3}{2}=\dfrac{8+27}{2}=\dfrac{35}{2}$

$\therefore\,B=\dfrac{35}{36}$

$\displaystyle\int{\dfrac{4{e}^{2x}+6}{9{e}^{2x}-4}dx}$

$=\displaystyle\int{\dfrac{A\left(9{e}^{2x}-4\right)+B\left(18{e}^{2x}\right)}{9{e}^{2x}-4}dx}$

$=\displaystyle\int{\dfrac{A\left(9{e}^{2x}-4\right)dx}{9{e}^{2x}-4}}+\displaystyle\int{\dfrac{B\left(18{e}^{2x}\right)dx}{9{e}^{2x}-4}}$

$=-\dfrac{3}{2}\displaystyle\int{\dfrac{\left(9{e}^{2x}-4\right)dx}{9{e}^{2x}-4}}+\dfrac{35}{36}\displaystyle\int{\dfrac{\left(18{e}^{2x}\right)dx}{9{e}^{2x}-4}}$

$=-\dfrac{3}{2}\displaystyle\int{dx}+\dfrac{35}{36}\displaystyle\int{\dfrac{\left(18{e}^{2x}\right)dx}{9{e}^{2x}-4}}$

$=-\dfrac{3}{2}x+\dfrac{35}{36}\log_{e}{\left(9{e}^{2x}-4\right)}+c$

Comparing with $Ax+B\log_{e}{\left(9{e}^{2x}-4\right)}+C$

we get $A=-\dfrac{3}{2},\,B=\dfrac{35}{36},\,C=c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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