What is the integral of 1/(x(10^{-1)-10^{-7}x)} ?

The integral of 1/(x(10^{-1)-10^{-7}x)} is -10(ln| 1/10-1/10000000 x|-ln(1/1000000 |x|))+C

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integral of 1/(x(10^{-1)-10^{-7}x)}

\int \frac{1}{x ( 1 0 ^{- 1} - 1 0 ^{- 7} x )} d x

- 10 (ln \frac{1}{10} - \frac{1}{10000000} x - ln (\frac{1}{1000000} ∣ x ∣)) + C

Solution steps

\int \frac{1}{x ( 1 0 ^{- 1} - 1 0 ^{- 7} x )} d x

Apply u-substitution

= \int - \frac{10}{u ( - 10 u + 1 )} d u

Take the constant out:

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

= - 10 \cdot \int \frac{1}{u ( - 10 u + 1 )} d u

Take the partial fraction of

\frac{1}{u ( - 10 u + 1 )} : \frac{1}{u} - \frac{10}{10 u - 1}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{10}{10 u - 1} d u = ln ∣ 10 u - 1 ∣

= - 10 (ln ∣ u ∣ - ln ∣ 10 u - 1 ∣)

Substitute back

u = 1 0^{- 1} - 1 0^{- 7} x

= - 10 (ln 1 0^{- 1} - 1 0^{- 7} x - ln 10 (1 0^{- 1} - 1 0^{- 7} x) - 1)

Simplify

- 10 (ln 1 0^{- 1} - 1 0^{- 7} x - ln 10 (1 0^{- 1} - 1 0^{- 7} x) - 1) : - 10 (ln \frac{1}{10} - \frac{1}{10000000} x - ln (\frac{1}{1000000} ∣ x ∣))

= - 10 (ln \frac{1}{10} - \frac{1}{10000000} x - ln (\frac{1}{1000000} ∣ x ∣))

Add a constant to the solution

= - 10 (ln \frac{1}{10} - \frac{1}{10000000} x - ln (\frac{1}{1000000} ∣ x ∣)) + C

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What is the integral of 1/(x(10^{-1)-10^{-7}x)} ?
The integral of 1/(x(10^{-1)-10^{-7}x)} is -10(ln| 1/10-1/10000000 x|-ln(1/1000000 |x|))+C