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

Question

Aicharya · Accepted Answer

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

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

Solution