Question

Let $f(x)=\int \frac{1+(x \ln x)^2}{x \ln x}dx$ such that $f(e)=\frac{e^2}{4}$. Then the value of $4\left(e^2f\left(\frac{1}{e}\right)+1\right)$ is

Answer 1

1

Answer 2

To find the value of $4\left(e^2f\left(\frac{1}{e}\right)+1\right)$, we first need to determine the function $f(x)$ by evaluating the given integral and then use the initial condition to find the constant of integration.

The given integral is $f(x)=\int \frac{1+(x \ln x)^2}{x \ln x}dx$. We can split the integrand into two parts: $f(x)=\int \left(\frac{1}{x \ln x} + \frac{(x \ln x)^2}{x \ln x}\right)dx$ $f(x)=\int \left(\frac{1}{x \ln x} + x \ln x\right)dx$ This can be written as the sum of two integrals: $f(x)=\int \frac{1}{x \ln x} dx + \int x \ln x dx$

Let's evaluate the first integral, $I_1 = \int \frac{1}{x \ln x} dx$: Let $u = \ln x$. Then, $du = \frac{1}{x} dx$. Substituting these into the integral: $I_1 = \int \frac{1}{u} du = \ln|u| + C_1 = \ln|\ln x| + C_1$.

Now, let's evaluate the second integral, $I_2 = \int x \ln x dx$: We use integration by parts, which states $\int p dq = pq - \int q dp$. Let $p = \ln x$ and $dq = x dx$. Then, $dp = \frac{1}{x} dx$ and $q = \frac{x^2}{2}$. Substituting these into the integration by parts formula: $I_2 = (\ln x)\left(\frac{x^2}{2}\right) - \int \left(\frac{x^2}{2}\right)\left(\frac{1}{x}\right) dx$ $I_2 = \frac{x^2}{2} \ln x - \int \frac{x}{2} dx$ $I_2 = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C_2$.

Combining the results for $I_1$ and $I_2$, we get the function $f(x)$: $f(x) = \ln|\ln x| + \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$, where $C = C_1 + C_2$.

Next, we use the given condition $f(e)=\frac{e^2}{4}$ to find the value of $C$. Substitute $x=e$ into the expression for $f(x)$: $f(e) = \ln|\ln e| + \frac{e^2}{2} \ln e - \frac{e^2}{4} + C$ Since $\ln e = 1$: $f(e) = \ln|1| + \frac{e^2}{2}(1) - \frac{e^2}{4} + C$ $f(e) = 0 + \frac{e^2}{2} - \frac{e^2}{4} + C$ $f(e) = \frac{2e^2 - e^2}{4} + C$ $f(e) = \frac{e^2}{4} + C$. We are given that $f(e)=\frac{e^2}{4}$. Therefore: $\frac{e^2}{4} = \frac{e^2}{4} + C$ This implies $C=0$.

So, the function $f(x)$ is: $f(x) = \ln|\ln x| + \frac{x^2}{2} \ln x - \frac{x^2}{4}$.

Finally, we need to find the value of $4\left(e^2f\left(\frac{1}{e}\right)+1\right)$. First, let's evaluate $f\left(\frac{1}{e}\right)$: Substitute $x=\frac{1}{e}$ into the expression for $f(x)$: $f\left(\frac{1}{e}\right) = \ln\left|\ln\left(\frac{1}{e}\right)\right| + \frac{\left(\frac{1}{e}\right)^2}{2} \ln\left(\frac{1}{e}\right) - \frac{\left(\frac{1}{e}\right)^2}{4}$ We know that $\ln\left(\frac{1}{e}\right) = \ln(e^{-1}) = -1$. So, $f\left(\frac{1}{e}\right) = \ln|-1| + \frac{1}{2e^2}(-1) - \frac{1}{4e^2}$ Since $\ln|-1| = \ln(1) = 0$: $f\left(\frac{1}{e}\right) = 0 - \frac{1}{2e^2} - \frac{1}{4e^2}$ $f\left(\frac{1}{e}\right) = -\frac{2}{4e^2} - \frac{1}{4e^2}$ $f\left(\frac{1}{e}\right) = -\frac{3}{4e^2}$.

Now, substitute this value into the expression $4\left(e^2f\left(\frac{1}{e}\right)+1\right)$: $4\left(e^2\left(-\frac{3}{4e^2}\right)+1\right)$ $= 4\left(-\frac{3}{4}+1\right)$ $= 4\left(\frac{-3+4}{4}\right)$ $= 4\left(\frac{1}{4}\right)$ $= 1$.

The final answer is $\boxed{1}$.

Explanation of the solution:

1. Decompose the integral: The integrand $\frac{1+(x \ln x)^2}{x \ln x}$ is split into two simpler terms: $\frac{1}{x \ln x}$ and $x \ln x$.
2. Integrate each term:
   • $\int \frac{1}{x \ln x} dx$: Use substitution $u = \ln x$, leading to $\ln|\ln x|$.
   • $\int x \ln x dx$: Use integration by parts, leading to $\frac{x^2}{2} \ln x - \frac{x^2}{4}$.
3. Combine and find constant: Sum the results of the two integrals to get $f(x) = \ln|\ln x| + \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$. Use the condition $f(e) = \frac{e^2}{4}$ to determine $C=0$.
4. Evaluate at specific point: Calculate $f\left(\frac{1}{e}\right)$ by substituting $x=\frac{1}{e}$ into the derived $f(x)$, which gives $-\frac{3}{4e^2}$.
5. Final calculation: Substitute $f\left(\frac{1}{e}\right)$ into the expression $4\left(e^2f\left(\frac{1}{e}\right)+1\right)$ and simplify to get 1.