If \(\int_{0}^{1} \frac{1}{\left(5+2 \mathrm{x}-2 \mathrm{x}^{2}\right)\left(1+\mathrm{e}^{(2-4 \mathrm{x})}\right)} \mathrm{dx}=\frac{1}{\alpha} \log _{\mathrm{e}}\left(\frac{\alpha+1}{\beta}\right)\), \(\alpha, \beta>0\), then \(\alpha^{4}-\beta^{4}\) is equal to:

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Definite Integrals

\int_{0}^{1} \frac{1}{( 5 + 2 x - 2 x ^{2} ) ( 1 + e ^{(2 - 4 x)} )} dx = \frac{1}{α} lo g_{e} (\frac{α + 1}{β})

α, β > 0

, then

α^{4} - β^{4}

is equal to:

Answer: (A)

Let

I

be the integral. The quadratic

5 + 2 x - 2 x^{2}

is unchanged under

x \to 1 - x

, while

e^{2 - 4 x} \to e^{- (2 - 4 x)}

, so replacing

x

1 - x

gives

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

. Adding the two forms,

2 I = \int_{0}^{1} \frac{d x}{5 + 2 x - 2 x ^{2}}

. Completing the square,

5 + 2 x - 2 x^{2} = 2 [\frac{11}{4} - (x - \frac{1}{2})^{2}]

, so

2 I = \frac{1}{11} lo g_{e} \frac{11 + 1}{11 - 1} = \frac{2}{11} lo g_{e} \frac{11 + 1}{10}

. Thus

I = \frac{1}{11} lo g_{e} \frac{11 + 1}{10}

, giving

α = 11, β = 10

, so

α^{4} - β^{4} = 121 - 100 = 21

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Single correct option