JEE Main 2023 — Definite Integration Question with Solution

1. The given integral is:

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

1. Perform a substitution, $x\to 1-x$. This gives:

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

1. Simplify the expression inside the integral:

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

1. Add the original integral (i) and the integral after substitution (ii):

$$2I={\int }_0^1\frac{dx}{5+2x-2x^2}$$

1. Factor the quadratic expression in the denominator:

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

1. To solve this integral, we can perform a change of variables using the substitution
   
   $x-\frac{1}{2}=\frac{\sqrt{11}}{2}\tan u$, then $dx=\frac{\sqrt{11}}{2}{\sec }^{2}udu$:

$$I=\frac{1}{\sqrt{11}}\int \frac{{\sec }^{2}u}{1+{\tan }^{2}u}du$$

1. Using the identity ${\sec }^{2}u=1+{\tan }^{2}u$:

$$I=\frac{1}{\sqrt{11}}\int du=\frac{1}{\sqrt{11}}(u+C)$$

1. Now we need to find the limits of the integral after the substitution. If $x=0$, then $u={\tan }^{-1}\left(-\frac{1}{\sqrt{11}}\right)$. If $x=1$, then $u={\tan }^{-1}\left(\frac{1}{\sqrt{11}}\right)$. So, the integral becomes:

$$I=\frac{1}{\sqrt{11}}\left[{\tan }^{-1}\left(\frac{1}{\sqrt{11}}\right)-{\tan }^{-1}\left(-\frac{1}{\sqrt{11}}\right)\right]$$

1. Using the properties of the arctangent function, we can rewrite the integral as:

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

1. From this result, we have $\alpha =\sqrt{11}$ and $\beta =\sqrt{10}$. Now, we can find ${\alpha }^4-{\beta }^4$:

$${\alpha }^4-{\beta }^4=(11{)}^2-(10{)}^2=121-100=21$$

Thus, ${\alpha }^4-{\beta }^4=21$.