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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

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

Simplify

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

= \frac{x}{e ^{x}} + \frac{e ^{x}}{x}

Add a constant to the solution

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

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