What is the integral of 1/(10r^{1/2)-r} ?

The integral of 1/(10r^{1/2)-r} is -2ln|10-r^{1/2}|+C

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integral of 1/(10r^{1/2)-r}

\int \frac{1}{10 r ^{\frac{1}{2}} - r} d r

- 2 ln 10 - r^{\frac{1}{2}} + C

Solution steps

\int \frac{1}{10 r ^{\frac{1}{2}} - r} d r

Apply u-substitution

: \int \frac{2}{10 - u} d u

= \int \frac{2}{10 - u} d u

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{1}{v} d v

= 2 \cdot \int - \frac{1}{v} d v

Take the constant out:

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

= 2 (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 (- ln ∣ v ∣)

Substitute back

= 2 (- ln 10 - r^{\frac{1}{2}})

Simplify

= - 2 ln 10 - r^{\frac{1}{2}}

Add a constant to the solution

= - 2 ln 10 - r^{\frac{1}{2}} + C

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What is the integral of 1/(10r^{1/2)-r} ?
The integral of 1/(10r^{1/2)-r} is -2ln|10-r^{1/2}|+C