What is the integral of (1+x)/(x^3)e^{1/x} ?

The integral of (1+x)/(x^3)e^{1/x} is -(e^{1/x})/x+C

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integral of (1+x)/(x^3)e^{1/x}

\int \frac{1 + x}{x ^{3}} e^{\frac{1}{x}} d x

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

Solution steps

\int \frac{1 + x}{x ^{3}} e^{\frac{1}{x}} d x

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \int \frac{e ^{\frac{1}{x}} ( 1 + x )}{x ^{3}} d x

Expand

\frac{e ^{\frac{1}{x}} ( 1 + x )}{x ^{3}} : \frac{e ^{\frac{1}{x}}}{x ^{3}} + \frac{e ^{\frac{1}{x}}}{x ^{2}}

= \int \frac{e ^{\frac{1}{x}}}{x ^{3}} + \frac{e ^{\frac{1}{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 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x + \int \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

\int \frac{e ^{\frac{1}{x}}}{x ^{3}} d x = - \frac{e ^{\frac{1}{x}}}{x} + e^{\frac{1}{x}}

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

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

Simplify

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

Add a constant to the solution

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

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What is the integral of (1+x)/(x^3)e^{1/x} ?
The integral of (1+x)/(x^3)e^{1/x} is -(e^{1/x})/x+C