What is the integral of 1/(x^2(e^{1/x)-5)} ?

The integral of 1/(x^2(e^{1/x)-5)} is 1/(5x)-1/5 ln|e^{1/x}-5|+C

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integral of 1/(x^2(e^{1/x)-5)}

\int \frac{1}{x ^{2} ( e ^{\frac{1}{x}} - 5 )} d x

\frac{1}{5 x} - \frac{1}{5} ln e^{\frac{1}{x}} - 5 + C

Solution steps

\int \frac{1}{x ^{2} ( e ^{\frac{1}{x}} - 5 )} d x

Apply u-substitution

: \int - \frac{1}{e ^{u} - 5} d u

= \int - \frac{1}{e ^{u} - 5} d u

Take the constant out:

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

= - \int \frac{1}{e ^{u} - 5} d u

Apply u-substitution

: \int \frac{1}{v ( v - 5 )} d v

= - \int \frac{1}{v ( v - 5 )} d v

Take the partial fraction of

\frac{1}{v ( v - 5 )} : - \frac{1}{5 v} + \frac{1}{5 ( v - 5 )}

= - \int - \frac{1}{5 v} + \frac{1}{5 ( v - 5 )} d v

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{1}{5 v} d v + \int \frac{1}{5 ( v - 5 )} d v)

\int \frac{1}{5 v} d v = \frac{1}{5} ln ∣ v ∣

\int \frac{1}{5 ( v - 5 )} d v = \frac{1}{5} ln ∣ v - 5 ∣

= - (- \frac{1}{5} ln ∣ v ∣ + \frac{1}{5} ln ∣ v - 5 ∣)

Substitute back

= - (- \frac{1}{5} ln e^{\frac{1}{x}} + \frac{1}{5} ln e^{\frac{1}{x}} - 5)

Simplify

- (- \frac{1}{5} ln e^{\frac{1}{x}} + \frac{1}{5} ln e^{\frac{1}{x}} - 5) : \frac{1}{5 x} - \frac{1}{5} ln e^{\frac{1}{x}} - 5

= \frac{1}{5 x} - \frac{1}{5} ln e^{\frac{1}{x}} - 5

Add a constant to the solution

= \frac{1}{5 x} - \frac{1}{5} ln e^{\frac{1}{x}} - 5 + C

\int \frac{1}{2} x^{2} - 2 x + 6 d x

\frac{3}{s} + \frac{24}{s ^{2}}

\frac{d ^{2} y}{d x} + 4 \frac{d y}{d x} - 12 y = 0

\frac{\partial}{\partial x} (x sin (8 x^{2} y))

\frac{- 60}{- 2 + 5 s}

What is the integral of 1/(x^2(e^{1/x)-5)} ?
The integral of 1/(x^2(e^{1/x)-5)} is 1/(5x)-1/5 ln|e^{1/x}-5|+C