Nguyên hàm 9

Tìm nguyên hàm. :$\displaystyle\int\left( x+1-\frac{1}{x} \right)e^{x+\frac{1}{x}}dx$

$\begin{aligned}\displaystyle\int\left( x+1-\frac{1}{x} \right)e^{x+\frac{1}{x}}dx&=\int e^{x+\frac{1}{x}}dx+\int x\left(1-\frac{1}{x^2}\right)e^{x+\frac{1}{x}}dx\\ &=\int e^{x+\frac{1}{x}}dx+\int xd\left(e^{x+\frac{1}{x}}\right)\\ &=\int e^{x+\frac{1}{x}}dx+x.e^{x+\frac{1}{x}}-\int e^{x+\frac{1}{x}}dx\\ &=x.e^{x+\frac{1}{x}}+C\end{aligned}$