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

The integral of x/(1+sqrt(x-1)) is 2/3 (x-1)^{3/2}-x+1+4sqrt(x-1)-4ln|sqrt(x-1)+1|+C

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

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

\frac{2}{3} (x - 1)^{\frac{3}{2}} - x + 1 + 4 x - 1 - 4 ln x - 1 + 1 + C

Solution steps

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

Apply u-substitution

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

= \int \frac{u + 1}{1 + u} d u

Apply u-substitution

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

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

Take the constant out:

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

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

Long division

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

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

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

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

\int 2 d v = 2 v

\int \frac{2}{v + 1} d v = 2 ln ∣ v + 1 ∣

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

Substitute back

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

Simplify

2 (\frac{( x - 1 ) ^{3}}{3} - \frac{( x - 1 ) ^{2}}{2} + 2 x - 1 - 2 ln x - 1 + 1) : \frac{2}{3} (x - 1)^{\frac{3}{2}} - x + 1 + 4 x - 1 - 4 ln x - 1 + 1

= \frac{2}{3} (x - 1)^{\frac{3}{2}} - x + 1 + 4 x - 1 - 4 ln x - 1 + 1

Add a constant to the solution

= \frac{2}{3} (x - 1)^{\frac{3}{2}} - x + 1 + 4 x - 1 - 4 ln x - 1 + 1 + C

f (x) = arctan (x)

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What is the integral of x/(1+sqrt(x-1)) ?
The integral of x/(1+sqrt(x-1)) is 2/3 (x-1)^{3/2}-x+1+4sqrt(x-1)-4ln|sqrt(x-1)+1|+C