What is the integral of (e^{4/x})/(x^2) ?

The integral of (e^{4/x})/(x^2) is -1/4 e^{4/x}+C

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integral of (e^{4/x})/(x^2)

\int \frac{e ^{\frac{4}{x}}}{x ^{2}} d x

- \frac{1}{4} e^{\frac{4}{x}} + C

Solution steps

\int \frac{e ^{\frac{4}{x}}}{x ^{2}} d x

Apply u-substitution

: \int - \frac{e ^{u}}{4} d u

= \int - \frac{e ^{u}}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{4} e^{u}

Substitute back

u = \frac{4}{x}

= - \frac{1}{4} e^{\frac{4}{x}}

Add a constant to the solution

= - \frac{1}{4} e^{\frac{4}{x}} + C

x \to - \infty lim (xx)

x \frac{d y}{d x} - (1 + x) y = y^{2}

\frac{s + 7}{( s + 2 ) ( s - 5 )}

y = \frac{e ^{x}}{ln ( x )}

What is the integral of (e^{4/x})/(x^2) ?
The integral of (e^{4/x})/(x^2) is -1/4 e^{4/x}+C