What is the integral of 1/(sqrt(x)(\sqrt{x)-1)} ?

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

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integral of 1/(sqrt(x)(\sqrt{x)-1)}

\int \frac{1}{x ( x - 1 )} d x

2 ln x - 1 + C

Solution steps

\int \frac{1}{x ( x - 1 )} 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 - 1

Add a constant to the solution

= 2 ln x - 1 + C

x y^{^{'}} + 2 y = 2 e x

\frac{d}{d x} (sec^{2} (3 x))

y^{^{''}} - 2 \cdot y^{^{'}} = sin (x)

\int_{0}^{24} 9 + 3 x d x

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 6)^{9})

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