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

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

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

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

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

Solution steps

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

Simplify

e^{\frac{1}{x}} \frac{1}{x ^{3}} : \frac{e ^{\frac{1}{x}}}{x ^{3}}

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

Apply u-substitution

: \int - e^{u} u d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= - (e^{u} u - \int e^{u} d u)

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

= - (e^{u} u - e^{u})

Substitute back

u = \frac{1}{x}

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} - 4 y^{^{'}} + 3 y = t^{2}

\frac{d}{d x} (\frac{e ^{- x} - x e ^{- x}}{e ^{x}})

e^{- 0.75 t} \cdot cos (4 t)

\frac{d}{d x} (3^{(1 - 3 x)})

20 + \frac{10}{s} \cdot \frac{1}{8 s + 1}

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