The integral [((x)/(2))^x+((2)/(x))^x] ((ex)/(2))dx is equal to :

Question

The integral $\int [{(\frac{x}{2})}^{x}+{(\frac{2}{x})}^{x}]\ln (\frac{ex}{2})dx$ is equal to :

Accepted Answer

To solve the integral:

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

we start by simplifying the logarithmic term:

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

So the integral becomes:

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

Let's denote:

$A={(\frac{x}{2})}^{x}$

$B={(\frac{2}{x})}^{x}$

Then, the integral can be split:

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

Calculating $\int A(1+\ln (\frac{x}{2}))\, dx$:
First, find the derivative of $A$:

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

This means:

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

Therefore:

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

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

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

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

The derivative of $B$ is:

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

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

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

Thus:

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

Therefore:

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

Combining the Results:
Adding both integrals:

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

Answer: Option B

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

Solutions · 1