Definite Integration Question - None

Question

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

Accepted Answer

Let ${I}=\int_{0}^{1} \frac{1}{\left(5+2 x-2 x^{2} ight)\left(1+e^{2-4 x} ight)} d x$ ...(i)
Replacing $x$ by $1-x$
$\Rightarrow {I}=\int_{0}^{1} \frac{1}{\left(5+2(1-x)-2(1-x)^{2} ight)\left(1+e^{2-4(1-x)} ight)} d x$
$=\int_{0}^{1} \frac{1}{\left(5+2-2 x-2-2 x^{2}+4 x ight)\left(1+e^{-2+4 x} ight)} d x$
$=\int_{0}^{1} \frac{1}{\left(5+2 x-2 x^{2} ight)\left(1+\frac{1}{e^{2-4 x}} ight)} d x$
${I}=\int_{0}^{1} \frac{e^{2-4 x}}{\left(5+2 x-2 x^{2} ight)\left(1+e^{2-4 x} ight)} d x$ ...(ii)
Adding (1) and (ii) we get,
$2 {I}=\int_{0}^{1} \frac{1}{5+2 x-2 x^{2}} d x=\int_{0}^{1} \frac{d x}{2\left(\frac{5}{2}+x-x^{2}+\frac{1}{4}-\frac{1}{4} ight)}$
$=\frac{1}{2} \int_{0}^{1} \frac{d x}{\left(\frac{\sqrt{11}}{2} ight)^{2}-\left(x-\frac{1}{2} ight)^{2}}=\frac{1 imes 2}{2 imes 2 imes \sqrt{11}}$ $\log \left|\frac{\frac{\sqrt{11}}{2}+x-\frac{1}{2}}{\frac{\sqrt{11}}{2}-x+\frac{1}{2}} ight|_{0}^{1}$
$=\frac{1}{2 \sqrt{11}}\left\{\log \left|\frac{\frac{\sqrt{11}}{2}+1-\frac{1}{2}}{\frac{\sqrt{11}}{2}-1+\frac{1}{2}} ight|-\log \left|\frac{\frac{\sqrt{11}}{2}-\frac{1}{2}}{\frac{\sqrt{11}}{2}+\frac{1}{2}} ight| ight\}$
$=\frac{1}{2 \sqrt{11}}\left\{\log \left|\frac{\sqrt{11}+1}{\sqrt{11}-1} ight|-\log \left|\frac{\sqrt{11}-1}{\sqrt{11}+1} ight| ight\}$
$=\frac{1}{2 \sqrt{11}} \log \left(\frac{\sqrt{11}+1}{\sqrt{11}-1} ight)^{2}$
$\Rightarrow {I}=\frac{1}{\sqrt{11}} \log \left(\frac{\sqrt{11}+1}{\sqrt{10}} ight)=\frac{1}{\alpha} \log \left(\frac{\alpha+1}{\beta} ight)$
$\Rightarrow \alpha=\sqrt{11}, \beta=\sqrt{10}$
Hence, $\alpha^{4}-\beta^{4}=121-100=21$

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