If 0 1 1 ( 5 + 2 x - 2 x 2 ) ( 1 + e ( 2 - 4 x ) ) d x = 1 log e ( + 1 ) , 0 , then 4 -4 is equal to [2023]

Question

$If \int_{0}^{1} \frac{1}{(5 + 2 x - 2 x^{2}) (1 + e^{(2 - 4 x)})} d x = \frac{1}{α} \log_{e} (\frac{α + 1}{β}),$ $α, β > 0, then α^{4} - β^{4} is equal to$ [2023]

Smart Achievers · Accepted Answer

(3)

Let $I = \int_{0}^{1} \frac{1}{(5 + 2 x - 2 x^{2}) (1 + e^{(2 - 4 x)})} d x$

$I = \int_{0}^{1} \frac{1}{(5 + 2 x (1 - x))} \cdot \frac{e^{4 x}}{(e^{4 x} + e^{2})} d x$ ...(i)

We know that, $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (b + a - x) d x$

$∴$ $I = \int_{0}^{1} \frac{1}{5 + 2 (1 - x) \cdot x} \cdot \frac{e^{4 (1 - x)}}{e^{4 (1 - x)} + e^{2}} d x$ ...(ii)

Adding (i) and (ii), we get

$\Rightarrow 2 I = \int_{0}^{1} \frac{1}{5 + 2 (1 - x) \cdot x} [\frac{e^{4 x} e^{4 (1 - x)} + e^{4 x} e^{2} + e^{4 x} e^{4 (1 - x)} + e^{4 (1 - x)} \cdot e^{2}}{(e^{4 (1 - x)} + e^{2}) (e^{4 x} + e^{2})}] d x$

$\Rightarrow 2 I = \int_{0}^{1} \frac{d x}{5 + 2 (1 - x) \cdot x} = \int_{0}^{1} \frac{d x}{\frac{11}{2} - 2 {(x - \frac{1}{2})}^{2}}$

$= \int_{0}^{1} \frac{d x}{(\frac{11}{4} - {(x - \frac{1}{2})}^{2})}$

$\Rightarrow I = \frac{1}{\sqrt{11}} \ln (\frac{\sqrt{11} + 1}{\sqrt{10}})$

$∴$ $α = \sqrt{11} and β = \sqrt{10} ∴ α^{4} - β^{4} = 121 - 100 = 21$

Solution

Q.
$If \int_{0}^{1} \frac{1}{(5 + 2 x - 2 x^{2}) (1 + e^{(2 - 4 x)})} d x = \frac{1}{α} \log_{e} (\frac{α + 1}{β}),$ $α, β > 0, then α^{4} - β^{4} is equal to$ [2023]

1 - 21

2 0

3 21

4 19

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