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

The integral of x/(sqrt(x)-1) is 3x+2ln|sqrt(x)-1|+2/3 (sqrt(x)-1)^3-3+C

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

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

3 x + 2 ln x - 1 + \frac{2}{3} (x - 1)^{3} - 3 + C

Solution steps

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

Apply u-substitution

: \int \frac{2 u ^{3}}{u - 1} d u

= \int \frac{2 u ^{3}}{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{u ^{3}}{u - 1} d u

Apply u-substitution

: \int \frac{( v + 1 ) ^{3}}{v} d v

= 2 \cdot \int \frac{( v + 1 ) ^{3}}{v} d v

Expand

\frac{( v + 1 ) ^{3}}{v} : v^{2} + 3 v + 3 + \frac{1}{v}

= 2 \cdot \int v^{2} + 3 v + 3 + \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

= 2 (\int v^{2} d v + \int 3 v d v + \int 3 d v + \int \frac{1}{v} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int 3 v d v = \frac{3 v ^{2}}{2}

\int 3 d v = 3 v

\int \frac{1}{v} d v = ln ∣ v ∣

= 2 (\frac{v ^{3}}{3} + \frac{3 v ^{2}}{2} + 3 v + ln ∣ v ∣)

Substitute back

= 2 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln x - 1)

Simplify

2 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln x - 1) : 3 x + 2 ln x - 1 + \frac{2}{3} (x - 1)^{3} - 3

= 3 x + 2 ln x - 1 + \frac{2}{3} (x - 1)^{3} - 3

Add a constant to the solution

= 3 x + 2 ln x - 1 + \frac{2}{3} (x - 1)^{3} - 3 + C

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What is the integral of x/(sqrt(x)-1) ?
The integral of x/(sqrt(x)-1) is 3x+2ln|sqrt(x)-1|+2/3 (sqrt(x)-1)^3-3+C