If $\int \frac{4 e^{x} + 6 e^{- x}}{9 e^{x} - 4 e^{- x}} d x = a x + b log (9 e^{2 x} + c) + d$ , then descending order of $a, b, c$ , is

a, b, c

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b, c, a

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b, a, c

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c, a, b

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Solution

The correct option is C $b, a, c$
We can express the numerator as,
$4 e^{x} + 6 e^{- x} = A (9 e^{x} - 4 e^{- x}) + B (9 e^{x} + 4 e^{- x})$
Comparing, the corresponding coefficients,
A = $\frac{- 21}{36}$ and $B = \frac{35}{36}$
Rewriting the given expression,
$\int \frac{A (9 e^{x} - 4 e^{- x})}{(9 e^{x} - 4 e^{- x})} d x + \int \frac{B (9 e^{x} + 4 e^{- x})}{(9 e^{x} - 4 e^{- x})} d x$
Let, $(9 e^{x} - 4 e^{- x}) = t$
differentiating both sides,
$(9 e^{x} + 4 e^{- x}) d x = d t$
Putting back the values,
$\int A d x + \int \frac{B d t}{t}$
$= A x + B ln t + k$
$= A x + B ln (9 e^{x} - 4 e^{- x}) + k$
Hence, $a = \frac{- 21}{36}$ and $B = \frac{35}{36} c = - 4$
So descending order is $b$ , $a$ , $c$

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