What is the integral of (7e^{-1/x})/(x^2) ?

The integral of (7e^{-1/x})/(x^2) is 7e^{-1/x}+C

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integral of (7e^{-1/x})/(x^2)

\int \frac{7 e ^{- \frac{1}{x}}}{x ^{2}} d x

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

Solution steps

\int \frac{7 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

= 7 \cdot \int \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

Apply u-substitution

: \int e^{u} d u

= 7 \cdot \int e^{u} d u

Use the common integral:

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

= 7 e^{u}

Substitute back

u = - \frac{1}{x}

= 7 e^{- \frac{1}{x}}

Add a constant to the solution

= 7 e^{- \frac{1}{x}} + C

\int \frac{1}{2 a t + t ^{2}} d t

\int \frac{1}{1 + 7 x} d x

\int 1 - \frac{x ^{2}}{4} d x

\int \frac{x - 4}{x ^{2} - 2 x + 3} d x

\int t^{2} + 3 d t

What is the integral of (7e^{-1/x})/(x^2) ?
The integral of (7e^{-1/x})/(x^2) is 7e^{-1/x}+C