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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - e^{u} d u

= \int - e^{u} d u

Take the constant out:

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

= - \int e^{u} d u

Use the common integral:

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

= - e^{u}

Substitute back

u = \frac{1}{x}

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

Add a constant to the solution

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

y^{^{'}} + y = - x^{4}

\frac{\partial}{\partial y} (x + \frac{y}{z})

\int 7 x e^{6 x} d x

y = 2 x^{2} + 1, (3, 19)

f (x) = (1 + 2 x)^{2}, at x = 4

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