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

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

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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

= \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

: - \frac{1}{u}

= - 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

5 - (\frac{8}{x ^{2}})

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x^{3} (a^{3} - x^{3})^{5}

x (1 + x)^{3}

\frac{( 5 x - 6 )}{( x ^{2} - 2 x + 2 )}

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