UPSC NDA 2 2022 Mathematics Solved Paper

Concept:Integrate $e^x (1 + \ln x + x \ln x)$ by splitting into simpler integrals and using integration by parts for the term containing $x \ln x$.Explanation:Given: $I = \int e^x (1 + \ln x + x \ln x) \, dx$.Split the integral: $I = \int e^x \, dx + \int e^x \ln x \, dx + \int e^x x \ln x \, dx$.The first integral is $\int e^x \, dx = e^x$.Now apply integration by parts to $\int e^x x \ln x \, dx$.Let $u = x \ln x$ and $dv = e^x \, dx$. Then $du = (1 + \ln x) \, dx$ and $v = e^x$.So $\int e^x x \ln x \, dx = x \ln x e^x - \int e^x (1 + \ln x) \, dx$.Expand the remaining integral: $\int e^x (1 + \ln x) \, dx = \int e^x \, dx + \int e^x \ln x \, dx = e^x + \int e^x \ln x \, dx$.Substitute back: $\int e^x x \ln x \, dx = x e^x \ln x - e^x - \int e^x \ln x \, dx$.Now plug all parts into $I$:$I = e^x + \int e^x \ln x \, dx + \left( x e^x \ln x - e^x - \int e^x \ln x \, dx \right)$.Cancel $e^x$ terms: $e^x - e^x = 0$.Cancel $\int e^x \ln x \, dx$ terms: $+ \int e^x \ln x \, dx - \int e^x \ln x \, dx = 0$.Thus $I = x e^x \ln x + c$.Answer:$x e^x \ln x + c$, which corresponds to option A.