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

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

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

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

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

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int e^{\frac{1}{x}} d x - \int e^{\frac{1}{x + 1}} d x

\int e^{\frac{1}{x}} d x = e^{\frac{1}{x}} x - Ei (\frac{1}{x})

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

= e^{\frac{1}{x}} x - Ei (\frac{1}{x}) - (e^{\frac{1}{x + 1}} x + e^{\frac{1}{x + 1}} - Ei (\frac{1}{x + 1}))

Simplify

= e^{\frac{1}{x}} x - Ei (\frac{1}{x}) - e^{\frac{1}{x + 1}} x - e^{\frac{1}{x + 1}} + Ei (\frac{1}{x + 1})

Add a constant to the solution

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

\frac{d}{d x} (ln (e^{x} - y))

\int x^{3} x^{2} - 1 d x

\frac{d}{d x} (\frac{1}{( x ^{2} - 1 ) ( x ^{2} + 3 x + 3 )})

\int sin^{3} (7 x) cos^{- 2} (7 x) d x

t y^{^{'}} = 2 y + 3 t^{3} cos (2 t), y (\frac{π}{4}) = 0

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