What is the integral of (8e^{-1/x})/(x^2) ?

The integral of (8e^{-1/x})/(x^2) is 8e^{-1/x}+C

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integral of (8e^{-1/x})/(x^2)

\int \frac{8 e ^{- \frac{1}{x}}}{x ^{2}} d x

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

Solution steps

\int \frac{8 e ^{- \frac{1}{x}}}{x ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

Apply u-substitution

: \int e^{u} d u

= 8 \cdot \int e^{u} d u

Use the common integral:

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

= 8 e^{u}

Substitute back

u = - \frac{1}{x}

= 8 e^{- \frac{1}{x}}

Add a constant to the solution

= 8 e^{- \frac{1}{x}} + C

- \frac{6}{( 1 - x ) ^{4}}

\frac{d ^{2}}{d x ^{2}} ((- 6 - 7 t)^{6})

\int_{1}^{4} t d t

y^{^{'}} = e^{y + 2 x}

\int_{0}^{1} (\frac{3 t ^{2}}{1 + t ^{3}}) d t

What is the integral of (8e^{-1/x})/(x^2) ?
The integral of (8e^{-1/x})/(x^2) is 8e^{-1/x}+C