Question

Solve $\int e^x-\frac{1}{x}(1+\frac{1}{x^2})dx$=${ke}^{x-\frac{1}{x}}$+$c$. Find k

Answer 1

$\int e^{x-\frac{1}{x}}(1+\frac{1}{x^2})dxt=e^{x-\frac{1}{x}}\Rightarrow \frac{dt}{dx}=\frac{d}{dx}\left(e^{x-\frac{1}{x}}\right)=e^{x-\frac{1}{x}}\times \frac{d}{dx}(x-\frac{1}{x})=e^{x-\frac{1}{x}}\times (1+\frac{1}{x^2})\Rightarrow \int dt=\int e^{x-\frac{1}{x}}(1+\frac{1}{x^2})dx\Rightarrow \int e^{x-\frac{1}{x}}(1+\frac{1}{x^2})dx=t+c=e^{x-\frac{1}{x}}+c$