\[\integral\_{}^{}{e^{x}(\log x + (1/x^{2}}))dx =] | MATH - INTEGRATION BY PARTS, INTEGRAL OF THE FORM | SolveeIt

$\int_{}^{}{e^{x}(\log x + (1/x^{2}}))dx =$

e^x log x + c

e^x (log x – 1/x) + c

e^x(log x + 1/x) + c

(e^x /x²) + c

Answer

e^x (log x – 1/x) + c

Explanation

= + $\int_{}^{}\frac{1}{x^{2}}$ e^x dx

= log x . e^x – + $\int_{}^{}{\frac{1}{x^{2}}e^{x}dx}$

= log x e^x – $\left( \frac{1}{x}e^{x} + \int_{}^{}{\frac{1}{x^{2}}e^{x}dx} \right)$ + $\int_{}^{}{\frac{1}{x^{2}}e^{x}dx}$

= log x e^x – $\frac{1}{x}$ e^x + c

Question: \[\int_{}^{}{e^{x}(\log x + (1/x^{2}}))dx =\]...