x^{2}* lne^{x^{2}}d

asked by guest
on Jan 17, 2025 at 11:47 am



You asked:

Evaluate the expression: \({x}^{2} \ln\left( {e}^{{x}^{2}} d \right)\)

MathBot Answer:

\[{x}^{2} \ln\left( {e}^{{x}^{2}} d \right) = x^{2} \log{\left(d e^{x^{2}} \right)}\]


\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).