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

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

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

integral

\frac{1}{( x - x ^{\frac{1}{2}} )}

2 ln x^{\frac{1}{2}} - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{u - 1} d u

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

Apply u-substitution

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

= 2 \cdot \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 x^{\frac{1}{2}} - 1

Add a constant to the solution

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

e^{- 5} x

x^{\frac{- 1}{2}}

x^{2} 1 + 4 x^{2}

\frac{1}{( x ( x ^{2} + 1 ) ^{2} )}

5 + x

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