What is the integral of 1/(sqrt(2x)-1) ?

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

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integral of 1/(sqrt(2x)-1)

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

2 x - 1 + ln 2 x - 1 + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{v + 1}{v} : 1 + \frac{1}{v}

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

Apply the Sum Rule:

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

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

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= \frac{1}{2} \cdot 2 (v + ln ∣ v ∣)

Substitute back

= \frac{1}{2} \cdot 2 (2 x - 1 + ln 2 x - 1)

Simplify

\frac{1}{2} \cdot 2 (2 x - 1 + ln 2 x - 1) : 2 x - 1 + ln 2 x - 1

= 2 x - 1 + ln 2 x - 1

Add a constant to the solution

= 2 x - 1 + ln 2 x - 1 + C

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