$\int\left(1+x-\frac{1}{x}\right) \mathrm{e}^{x+\frac{1}{x}} \mathrm{~d} x$ equal to

Question

Aicharya · Accepted Answer

Note that $\int\left[x \mathrm{f}^{\prime}(x)+\mathrm{f}(x)ight] \mathrm{d} x=x \mathrm{f}(x)+\mathrm{c}$ $\begin{aligned} 	herefore \quad & \int\left(1+x-\frac{1}{x}ight) \mathrm{e}^{x+\frac{1}{x}} \mathrm{~d} x \ & =\int\left[x \mathrm{e}^{x+\frac{1}{x}}\left(1-\frac{1}{x^2}ight)+\mathrm{e}^{x+\frac{1}{x}}ight] \mathrm{d} x \ & =x \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c} \end{aligned}$

$\int\left(1+x-\frac{1}{x}\right) \mathrm{e}^{x+\frac{1}{x}} \mathrm{~d} x$ equal to

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