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

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

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

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

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

Solution steps

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

x \to \infty lim (∣ 2 - x ∣)

\frac{\partial}{\partial x} (cos (8 y - 3 x))

\frac{d}{d x} (\frac{1}{2} (cos (5 x))^{2})

y^{^{''}} - 2 y^{^{'}} = 12 x - 10

3 x^{2} + 2 x

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