What is the integral of 1/(sqrt(y)-1) ?

The integral of 1/(sqrt(y)-1) is 2(sqrt(y)-1+ln|sqrt(y)-1|)+C

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integral of 1/(sqrt(y)-1)

\int \frac{1}{y - 1} d y

2 (y - 1 + ln ∣ y - 1 ∣) + C

Solution steps

\int \frac{1}{y - 1} d y

Apply u-substitution

: \int \frac{2 ( u + 1 )}{u} d u

= \int \frac{2 ( u + 1 )}{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{u + 1}{u} d u

Expand

\frac{u + 1}{u} : 1 + \frac{1}{u}

= 2 \cdot \int 1 + \frac{1}{u} 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

= 2 (\int 1 d u + \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 (u + ln ∣ u ∣)

Substitute back

u = y - 1

= 2 (y - 1 + ln ∣ y - 1 ∣)

Add a constant to the solution

= 2 (y - 1 + ln ∣ y - 1 ∣) + C

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What is the integral of 1/(sqrt(y)-1) ?
The integral of 1/(sqrt(y)-1) is 2(sqrt(y)-1+ln|sqrt(y)-1|)+C