$\int \frac{e^{x}}{x} (x \cdot (log x)^{2} + 2 log x) d x$

Open in App

Solution

$\int \frac{e^{x}}{x} (x {(log x)}^{2} + 2 log x) d x = \int \frac{e^{x}}{x} (x {(log x)}^{2} + \frac{2 log x}{x}) d x = \int \frac{e^{x}}{x} (x {(log x)}^{2}) + \int \frac{e^{x}}{x} \frac{2 log x}{x} d x = e^{x} {(log x)}^{2} - \int e^{x} \frac{2 log x}{x} d x + \int e^{x} \frac{2 log x}{x} d x + C = e^{x} {(log x)}^{2} + C$

Suggest Corrections

Similar questions

\int \{\tan (\log x) + \sec^{2} (\log x)\} d x

\int_{0}^{\frac{π}{2}} log cos x d x = \frac{π}{2} log (\frac{1}{2})

\int_{0}^{π / 2} log sec x d x =

Join BYJU'S Learning Program