Question:

Step 1: Rewrite the integral using the differential $d(x^2)$. The integral is given by: $$I_n = \int_1^e (\ln x)^n d(x^2)$$ Step 2: Apply integration by parts. Let $u = (\ln x)^n$ and $dv = d(x^2)$. Then, the differential of $u$ is $du = n(\ln x)^{n-1} \cdot \frac{1}{x} dx$, and the integral of $dv$ is $v = x^2$. Using the integration by parts formula $\int u \, dv = uv - \int v \, du$: $$I_n = [x^2 (\ln x)^n]_1^e - \int_1^e x^2 \cdot n(\ln x)^{n-1} \cdot \frac{1}{x} dx$$ Step 3: Evaluate the boundary term and simplify the integral. First, evaluate the boundary term $[x^2 (\ln x)^n]_1^e$: At $x = e$: $e^2 (\ln e)^n = e^2 \cdot 1^n = e^2$. At $x = 1$: $1^2 (\ln 1)^n = 1 \cdot 0^n = 0$. So, the boundary term evaluates to $e^2 - 0 = e^2$. Next, simplify the integral term: $$n \int_1^e x^2 \cdot (\ln x)^{n-1} \cdot \frac{1}{x} dx = n \int_1^e x (\ln x)^{n-1} dx$$ Substituting these back into the expression for $I_n$: $$I_n = e^2 - n \int_1^e x (\ln x)^{n-1} dx$$ Step 4: Express the remaining integral in terms of $I_{n-1}$. The definition of $I_{n-1}$ is $I_{n-1} = \int_1^e (\ln x)^{n-1} d(x^2)$. Since $d(x^2) = 2x \, dx$, we can rewrite $I_{n-1}$ as: $$I_{n-1} = \int_1^e (\ln x)^{n-1} (2x \, dx) = 2 \int_1^e x (\ln x)^{n-1} dx$$ From this, we can express the integral term from Step 3 as: $$\int_1^e x (\ln x)^{n-1} dx = \frac{1}{2} I_{n-1}$$ Substitute this back into the equation for $I_n$: $$I_n = e^2 - n \left( \frac{1}{2} I_{n-1} \right)$$ $$I_n = e^2 - \frac{n}{2} I_{n-1}$$ Step 5: Rearrange the equation to find $2I_n + nI_{n-1}$. Multiply the equation $I_n = e^2 - \frac{n}{2} I_{n-1}$ by 2: $$2I_n = 2e^2 - nI_{n-1}$$ Add $nI_{n-1}$ to both sides of the equation: $$2I_n + nI_{n-1} = 2e^2$$