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

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

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integral of e^{x-1}+x^{-2}

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

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

Solution steps

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

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

\int x^{- 2} d x = - \frac{1}{x}

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

Add a constant to the solution

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

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

\frac{d}{d x} (\frac{x - cos ( x )}{x ^{2} + cot ( x )})

\frac{d}{d x} (\frac{9 x ^{2}}{x})

\frac{d y}{d x} - \frac{y}{x} = 5 x e^{x}

\int \frac{2}{- sin ( t ) + 1} d t

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