JEE Main 2023 — Definite Integration Question with Solution

Given,

${\int }_0^1\frac{1}{\left(5+2x-2x^2\right)\left(1+e^{\left(2-4x\right)}\right)}dx=\frac{1}{\alpha }{\log }_e\left(\frac{\alpha +1}{\beta }\right)$

Now let,

$I=\frac{1}{2}{\int }_0^1\frac{1}{\left(\frac{5}{2}-\left(x^2-x\right)\right)\left(1+e^{-4\left(x-\frac{1}{2}\right)}\right)}dx$

$\Rightarrow I=\frac{1}{2}{\int }_0^1\frac{1}{\left(\left(\frac{11}{4}\right)-{\left(x-\frac{1}{2}\right)}^2\right)\left(1+e^{-4\left(x-\frac{1}{2}\right)}\right)}dx$

Now let $x-\frac{1}{2}=t\Rightarrow dx=dt$

So, integral becomes,

$I=\frac{1}{2}{\int }_{\frac{-1}{2}}^{\frac{1}{2}}\frac{1}{\left.{\left(\frac{\sqrt{11}}{2}\right)}^2-t^2\right)\left(1+e^{-4t}\right)}dt$

$\Rightarrow I=\frac{1}{2}{\int }_{\frac{-1}{2}}^0\frac{1}{\left.{\left(\frac{\sqrt{11}}{2}\right)}^2-t^2\right)\left(1+e^{-4t}\right)}dt+\underset{I_2}{\underbrace{\frac{1}{2}{\int }_0^{\frac{1}{2}}\frac{1}{\left.{\left(\frac{\sqrt{11}}{2}\right)}^2-t^2\right)\left(1+e^{-4t}\right)}dt}}$

Now solving $I_2=\frac{1}{2}{\int }_{\frac{-1}{2}}^{\frac{1}{2}}\frac{1}{\left.{\left(\frac{\sqrt{11}}{2}\right)}^2-t^2\right)\left(1+e^{-4t}\right)}dt$, 

Let $t=-z$, we get $dt=-dz$

So, $I_2=-\frac{1}{2}{\int }_{\frac{1}{2}}^0\frac{dz}{\left.{\left(\frac{\sqrt{11}}{2}\right)}^2-z^2\right)\left(1+e^{4z}\right)}$

$\Rightarrow I_2=\frac{1}{2}{\int }_0^{\frac{1}{2}}\frac{dt}{\left.{\left(\frac{\sqrt{11}}{2}\right)}^2-t^2\right)\left(1+e^{4t}\right)}$

Now putting the value of $I_2 \text{in} I$ we get,

$I=\frac{1}{2}{\int }_0^{\frac{1}{2}}\frac{1}{\left({\left(\frac{\sqrt{11}}{2}\right)}^2-t^2\right)\left(1+e^{-4t}\right)}+\frac{1}{\left({\left(\frac{\sqrt{11}}{2}\right)}^2-t^2\right)\left(1+e^{4t}\right)}dt$

$\Rightarrow I=\frac{1}{2}{\int }_0^{\frac{1}{2}}\frac{e^{4t}}{\left({\left(\frac{\sqrt{11}}{2}\right)}^2-t^2\right)\left(e^{4t}+1\right)}+\frac{1}{\left({\left(\frac{\sqrt{11}}{2}\right)}^2-t^2\right)\left(1+e^{4t}\right)}dt$

$\Rightarrow I=\frac{1}{2}{\int }_0^{\frac{1}{2}}\frac{1}{{\left(\frac{\sqrt{11}}{2}\right)}^2-t^2}dt$

$\Rightarrow I=\frac{1}{2}\times \frac{1}{2\left(\frac{\sqrt{11}}{2}\right)}{\left[\ln \left|\frac{\frac{\sqrt{11}}{2}+t}{\frac{\sqrt{11}}{2}-t}\right|\right]}_0^{\frac{1}{2}}$

$\Rightarrow I=\frac{1}{2\sqrt{11}}\ln \left(\frac{\sqrt{11}+1}{\sqrt{11}-1}\right)=\frac{1}{2\sqrt{11}}\ln \left(\frac{(\sqrt{11}+1{)}^2}{10}\right)$

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

Now on comparing with $\frac{1}{\alpha }{\log }_e\left(\frac{\alpha +1}{\beta }\right)$ we get,

$\Rightarrow \alpha =\sqrt{11}, \beta =\sqrt{10}\Rightarrow {\alpha }^4-{\beta }^4=21$