Mathematics

If $\int { \dfrac { 4{ e }^{ x }+6{ e }^{ -x } }{ 9{ e }^{ x }-4{ e }^{ -x } } =Ax+B\log _{ e }{ \left( 9{ e }^{ 2x }-4 \right) +C } }$, then the value of $A, B$ and $C$ are

$\dfrac {-3}{2},\dfrac {35}{36}$ and $C$ is a constant

SOLUTION
$\int\dfrac{4x^x+6e^{-x}}{9e^x-4e^{-x}}=Ax+ B\ log_e(9e^{2x}-4)+ C$

we have to calculate the value of A, B, and C

$\text{If L.H.S and R.H.S are equal then differentiation of both side also must be equal.}$

so,

$\text{diff. of LHS= Diff of RHS}$

$\Rightarrow\dfrac{4e^x+6^{-x}}{9e^x-4e^{-x}}=A +B\dfrac{1}{(9e^{2x}-4)}.18e^{2x}$

$\Rightarrow\dfrac{4e^x+6^{-x}}{9e^x-4e^{-x}}=\dfrac{9Ae^{2x}- 4A+ 18Be^{2x}}{9e^{2x}- 4}$

take $e^{-x}$  comman from R.H.S

$\Rightarrow\dfrac{4e^x+6^{-x}}{9e^x-4e^{-x}}=\dfrac{9Ae^{x}- 4A^{-x}+ 18Be^{x}}{9e^{x}- 4e^{-x}}$

solve the above equation

$\Rightarrow 4e^x +6e^{-x}=9Ae^x-4Ae^{-x}+18Be^x$

Now, Compare both sides

$6 = -4A\Rightarrow A = -\dfrac32.........(i)$ and

$4 =9A+18B........(ii)$

from $(i) and (ii)$

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

$\therefore \text{option A is correct}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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