The integral [ [((x)/(2))^x+((2)/(x))^x] ((e x)/(2)) d x ] is equal to :

Question

The integral $$
\int\left[\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right] \ln \left(\frac{e x}{2}\right) d x
$$ is equal to :

Accepted Answer

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 $

Solutions · 1