What is the integral of (1-1/(x^2))e^{x+1/x} ?

The integral of (1-1/(x^2))e^{x+1/x} is e^{x+1/x}+C

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integral of (1-1/(x^2))e^{x+1/x}

\int (1 - \frac{1}{x ^{2}}) e^{x + \frac{1}{x}} d x

e^{x + \frac{1}{x}} + C

Solution steps

\int (1 - \frac{1}{x ^{2}}) e^{x + \frac{1}{x}} d x

Apply u-substitution

: \int e^{u} d u

= \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= e^{u}

Substitute back

u = x + \frac{1}{x}

= e^{x + \frac{1}{x}}

Add a constant to the solution

= e^{x + \frac{1}{x}} + C

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What is the integral of (1-1/(x^2))e^{x+1/x} ?
The integral of (1-1/(x^2))e^{x+1/x} is e^{x+1/x}+C