If _0^1dx(5+2x-2x^2)(1+e^2-4x) = 1_e(+1), ,0, then ^4-^4 is equal to:

Question

If $\displaystyle\int_0^1\frac{dx}{(5+2x-2x^2)\big(1+e^{2-4x}\big)} = \dfrac1\alpha\log_e\left(\dfrac{\alpha+1}\beta\right)$, $\alpha,\beta>0$, then $\alpha^4-\beta^4$ is equal to:

Accepted Answer

By King's rule ($x\to1-x$), $\dfrac1{1+e^{2-4x}}+\dfrac1{1+e^{2-4(1-x)}}=\dfrac1{1+e^{2-4x}}+\dfrac{e^{2-4x}}{1+e^{2-4x}}=1$. Adding the two forms:

$$2I = \int_0^1\frac{dx}{5+2x-2x^2} = \int_0^1\frac{dx}{2\left(\frac{11}4-\left(x-\frac12\right)^2\right)}$$

$$I = \frac1{\sqrt{11}}\ln\left(\frac{\sqrt{11}+1}{\sqrt{10}}\right)$$

So $\alpha=\sqrt{11}$, $\beta=\sqrt{10}$, giving $\alpha^4-\beta^4 = 121-100=21$.

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