If ∫01 15+2x−2x21+e(2−4x)dx=1αloge⁡α+1β,α,β>0 then α4−β4 is equal to

I=∫01 15+2x−2x21+e(2−4x)dx =12∫01 152−x2−x1+e−4x−12dx=12∫01 1114−x−1221+e−4x−12dx Let x−12=t⇒dx=dt=12∫−1212 11122−t21+e−4tdt=12∫012 11122−t21+e−4t+11122−t21+e4tdt=12∫012 11122−t2dt=12×12112ln⁡112+t112−t012=1211ln⁡11+111−1=1211ln⁡(11+1)210=111ln⁡11+110⇒α=11,β=10⇒α4−β4=21

If ∫01 15+2x−2x21+e(2−4x)dx=1αloge⁡α+1β,α,β>0 then α4−β4 is equal to

I=∫01 15+2x−2x21+e(2−4x)dx =12∫01 152−x2−x1+e−4x−12dx=12∫01 1114−x−1221+e−4x−12dx Let x−12=t⇒dx=dt=12∫−1212 11122−t21+e−4tdt=12∫012 11122−t21+e−4t+11122−t21+e4tdt=12∫012 11122−t2dt=12×12112ln⁡112+t112−t012=1211ln⁡11+111−1=1211ln⁡(11+1)210=111ln⁡11+110⇒α=11,β=10⇒α4−β4=21

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

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Detailed Solution

$\begin{array}{l} I = \int_{0}^{1} \frac{1}{(5 + 2 x - 2 x^{2}) (1 + e^{(2 - 4 x)})} dx = \frac{1}{2} \int_{0}^{1} \frac{1}{(\frac{5}{2} - (x^{2} - x)) (1 + e^{- 4 (x - \frac{1}{2})})} d x \\ = \frac{1}{2} \int_{0}^{1} \frac{1}{((\frac{11}{4}) - {(x - \frac{1}{2})}^{2}) (1 + e^{- 4 (x - \frac{1}{2})})} d x \end{array} Let x - \frac{1}{2} = t \Rightarrow d x = d t$
$\begin{array}{l} = \frac{1}{2} \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{1}{({(\frac{\sqrt{11}}{2})}^{2} - t^{2}) (1 + e^{- 4 t})} d t \\ = \frac{1}{2} \int_{0}^{\frac{1}{2}} [\frac{1}{({(\frac{\sqrt{11}}{2})}^{2} - t^{2}) (1 + e^{- 4 t})} + \frac{1}{({(\frac{\sqrt{11}}{2})}^{2} - t^{2}) (1 + e^{4 t})}] d t \\ = \frac{1}{2} \int_{0}^{\frac{1}{2}} \frac{1}{{(\frac{\sqrt{11}}{2})}^{2} - t^{2}} d t = \frac{1}{2} \times {\frac{1}{2 (\frac{\sqrt{11}}{2})} \ln |\frac{\frac{\sqrt{11}}{2} + t}{\frac{\sqrt{11}}{2} - t}||}_{0}^{\frac{1}{2}} \end{array}$
$\begin{array}{l} = \frac{1}{2 \sqrt{11}} \ln (\frac{\sqrt{11} + 1}{\sqrt{11} - 1}) = \frac{1}{2 \sqrt{11}} \ln (\frac{(\sqrt{11} + 1)^{2}}{10}) \\ = \frac{1}{\sqrt{11}} \ln (\frac{\sqrt{11} + 1}{\sqrt{10}}) \\ \Rightarrow α = \sqrt{11}, β = \sqrt{10} \Rightarrow α^{4} - β^{4} = 21 \end{array}$