What is the integral of (1-0.1x)^{-2} ?

The integral of (1-0.1x)^{-2} is (10)/(1-0.1x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (1-0.1x)^{-2}

integral

(1 - 0.1 x)^{- 2}

\frac{10}{1 - 0.1 x} + C

Solution steps

\int (1 - 0.1 x)^{- 2} d x

Apply u-substitution

= \int - \frac{1}{0.1 u ^{2}} d u

Take the constant out:

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

= - \frac{1}{0.1} \cdot \int \frac{1}{u ^{2}} d u

Simplify

= - 10 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

= - 10 (- \frac{1}{u})

Substitute back

u = 1 - 0.1 x

= - 10 (- \frac{1}{1 - 0.1 x})

Simplify

- 10 (- \frac{1}{1 - 0.1 x}) : \frac{10}{1 - 0.1 x}

= \frac{10}{1 - 0.1 x}

Add a constant to the solution

= \frac{10}{1 - 0.1 x} + C

What is the integral of (1-0.1x)^{-2} ?
The integral of (1-0.1x)^{-2} is (10)/(1-0.1x)+C