The integral \int \left(\right. 1 + x - \frac{1}{x} \left.\right) \textrm{ } e^{x + \frac{1}{x}} \textrm{ } dx is equal to

Engineering

Mathematics

Integration by Parts

Question

The integral $\int (1 + x - \frac{1}{x}) e^{x + \frac{1}{x}} dx$ is equal to

$- x e^{(x + \frac{1}{x})} + C$

$(x + 1) e^{(x + \frac{1}{x})} + C$

$x e^{(x + \frac{1}{x})} + C$

$(x - 1) e^{(x + \frac{1}{x})} + C$

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We know that $\int (f (x) + xf' (x)) dx$ = x f(x)

So, $\int (e^{(x + \frac{1}{x})} + (x - \frac{1}{x}) e^{(x + \frac{1}{x})}) dx$

Now, f(x) = $e^{(x + \frac{1}{x})}$

$∴$ x f(x) = x $e^{(x + \frac{1}{x})}$ .