Question

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

• A. $\frac{-3}{2},\frac{35}{36}$ and $C$ is a constant
• B. $-\frac{1}{2},\frac{3}{2}$ and $C=0$
• C. $\frac{-3}{2},\frac{33}{2}$ and $C=1$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{-3}{2},\frac{35}{36}$ and $C$ is a constant

$\int \frac{4x^x+6e^{-x}}{9e^x-4e^{-x}}=Ax+Blo{g}_e(9e^{2x}-4)+C$

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,

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

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

$\Rightarrow \frac{4e^x+6^{-x}}{9e^x-4e^{-x}}=\frac{9A{e}^{2x}-4A+18B{e}^{2x}}{9e^{2x}-4}$

take $e^{-x}$ comman from R.H.S

$\Rightarrow \frac{4e^x+6^{-x}}{9e^x-4e^{-x}}=\frac{9A{e}^x-4A^{-x}+18B{e}^x}{9e^x-4e^{-x}}$

solve the above equation

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

Now, Compare both sides

$6=-4A\Rightarrow A=-\frac{3}{2}.........\left(i\right)$ and

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

from $\left(i\right)and\left(ii\right)$

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

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