What is the integral of (9e^{-1/x})/(x^2) ?

The integral of (9e^{-1/x})/(x^2) is 9e^{-1/x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (9e^{-1/x})/(x^2)

\int \frac{9 e ^{- \frac{1}{x}}}{x ^{2}} d x

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

Solution steps

\int \frac{9 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

= 9 \cdot \int \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

Apply u-substitution

: \int e^{u} d u

= 9 \cdot \int e^{u} d u

Use the common integral:

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

= 9 e^{u}

Substitute back

u = - \frac{1}{x}

= 9 e^{- \frac{1}{x}}

Add a constant to the solution

= 9 e^{- \frac{1}{x}} + C

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

x^{4} - 4 x^{2} + 10, at x = 1

y^{^{'}} + x y = 0

x \to 0 + lim (cot^{2} (x))

\frac{d}{d x} (x^{4} - 9 x + 6)

What is the integral of (9e^{-1/x})/(x^2) ?
The integral of (9e^{-1/x})/(x^2) is 9e^{-1/x}+C