UPSC NDA I 2018 Mathematics Solved Paper

Concept:Integration by parts is used for products of functions. The formula is: $\int u \, dv = uv - \int v \, du$. The ILATE rule gives priority order: Inverse, Logarithmic, Algebraic, Trigonometric, Exponential. Standard integrals: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ and $\frac{d}{dx}(\ln x) = \frac{1}{x}$.Explanation:Let $I = \int_{1}^{e} x \ln x \, dx$. Use integration by parts. Take $u = \ln x$ and $dv = x \, dx$. Then $du = \frac{1}{x} \, dx$ and $v = \frac{x^2}{2}$. Applying the formula: $I = \left[ \ln x \cdot \frac{x^2}{2} \right]_{1}^{e} - \int_{1}^{e} \frac{1}{x} \cdot \frac{x^2}{2} \, dx$. Simplify: $I = \left[ \frac{x^2}{2} \ln x \right]_{1}^{e} - \frac{1}{2} \int_{1}^{e} x \, dx$. Evaluate the first term: at $x = e$, $\frac{e^2}{2} (\ln e) = \frac{e^2}{2} \cdot 1 = \frac{e^2}{2}$; at $x = 1$, $\frac{1^2}{2} (\ln 1) = \frac{1}{2} \cdot 0 = 0$. So $\left[ \frac{x^2}{2} \ln x \right]_{1}^{e} = \frac{e^2}{2} - 0 = \frac{e^2}{2}$. Now $\int_{1}^{e} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{e} = \frac{e^2}{2} - \frac{1}{2}$. Thus $I = \frac{e^2}{2} - \frac{1}{2} \left( \frac{e^2}{2} - \frac{1}{2} \right) = \frac{e^2}{2} - \frac{e^2}{4} + \frac{1}{4} = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2 + 1}{4}$.Answer:$\frac{e^{2} + 1}{4}$ (Option B).