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

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

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

integral

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

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

Solution steps

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

Simplify

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

3 x^{2} + 6 x - 2

(2^{x})^{2}

y = 2 x

e^{\frac{- x ^{2}}{4}}

\frac{( ( 1 + x ) ^{3} )}{( x )}

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