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


ANSWER

$$\dfrac {-3}{2},\dfrac {35}{36}$$ 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
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