QJEEM23Ma73cc: JEE Main 2023 Mathematics - Class 12 JEEM23Ma73cc Solution: The integral ∫ [ ( x 2 ) x + ( 2 x ) x ] ln ⁡ ( e x 2 ) d x is…

To solve the integral:

$I = \int\left[\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right] \ln \left(\frac{e x}{2}\right) \, dx$

we start by simplifying the logarithmic term:

$\ln\left(\frac{e x}{2}\right) = \ln(e) + \ln\left(\frac{x}{2}\right) = 1 + \ln\left(\frac{x}{2}\right)$

So the integral becomes:

$I = \int \left[ \left( \frac{x}{2} \right)^x + \left( \frac{2}{x} \right)^x \right] \left[ 1 + \ln\left( \frac{x}{2} \right) \right] \, dx$

Let's denote:

$A = \left( \dfrac{x}{2} \right)^x$

$B = \left( \dfrac{2}{x} \right)^x$

Then, the integral can be split:

$I = \int \left[ A(1 + \ln\left( \frac{x}{2} \right)) + B(1 + \ln\left( \frac{x}{2} \right)) \right] \, dx$

Calculating $\int A(1 + \ln\left( \frac{x}{2} \right)) \, dx$ :

First, find the derivative of $A$ :

$\frac{dA}{dx} = A \left[ \ln\left( \frac{x}{2} \right) + 1 \right]$

This means:

$A \left[ 1 + \ln\left( \frac{x}{2} \right) \right] = \frac{dA}{dx}$

Therefore:

$\int A \left[ 1 + \ln\left( \frac{x}{2} \right) \right] \, dx = \int \frac{dA}{dx} \, dx = A + C_1 = \left( \frac{x}{2} \right)^x + C_1$

Calculating $\int B(1 + \ln\left( \frac{x}{2} \right)) \, dx$ :

Note that:

$\ln\left( \frac{x}{2} \right) = \ln x - \ln 2$

$\ln\left( \frac{2}{x} \right) = \ln 2 - \ln x$

The derivative of $B$ is:

$\frac{dB}{dx} = B \left[ \ln\left( \frac{2}{x} \right) + 1 \right] = B \left[ \ln 2 - \ln x + 1 \right]$

But in our integrand, we have $B \left[ 1 + \ln\left( \frac{x}{2} \right) \right] = B \left[ 1 + \ln x - \ln 2 \right]$ . Notice that:

$1 + \ln x - \ln 2 = - \left( \ln 2 - \ln x + 1 \right)$

Thus:

$B \left[ 1 + \ln\left( \frac{x}{2} \right) \right] = - \frac{dB}{dx}$

Therefore:

$\int B \left[ 1 + \ln\left( \frac{x}{2} \right) \right] \, dx = -\int \frac{dB}{dx} \, dx = -B + C_2 = -\left( \frac{2}{x} \right)^x + C_2$

Combining the Results:

Adding both integrals:

$I = \left( \frac{x}{2} \right)^x - \left( \frac{2}{x} \right)^x + C$

Answer: Option B

$\left( \frac{x}{2} \right)^x - \left( \frac{2}{x} \right)^x + C$