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

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

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

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

- e^{x} x^{2} + 2 e^{x} x - e^{x} + C

Solution steps

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

Simplify

\frac{( x ^{- 2} - 1 ) e ^{x}}{x ^{- 2}} : e^{x} - e^{x} x^{2}

= \int e^{x} - e^{x} 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} d x - \int e^{x} x^{2} d x

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

\int e^{x} x^{2} d x = x^{2} e^{x} - 2 (e^{x} x - e^{x})

= e^{x} - (x^{2} e^{x} - 2 (e^{x} x - e^{x}))

Simplify

e^{x} - (x^{2} e^{x} - 2 (e^{x} x - e^{x})) : - e^{x} x^{2} + 2 e^{x} x - e^{x}

= - e^{x} x^{2} + 2 e^{x} x - e^{x}

Add a constant to the solution

= - e^{x} x^{2} + 2 e^{x} x - e^{x} + C

\int \frac{1}{x ^{2} \pm a ^{2}} d x

\int \frac{0.3}{r ln ( r )} d r

\int \frac{1}{u u ^{2} - 1} d u

\int \frac{( 1 - ln ( x ) )}{x} d x

\int \frac{d}{1} (\frac{ln ( x )}{x}) d x

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