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

The integral of (sqrt(u-1))/u is 2(-arctan(sqrt(u-1))+sqrt(u-1))+C

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

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

2 (- arctan (u - 1) + u - 1) + C

Solution steps

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

Apply u-substitution

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

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

\frac{v ^{2}}{v ^{2} + 1} = - \frac{1}{v ^{2} + 1} + 1

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

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

\int 1 d v = v

= 2 (- arctan (v) + v)

Substitute back

v = u - 1

= 2 (- arctan (u - 1) + u - 1)

Add a constant to the solution

= 2 (- arctan (u - 1) + u - 1) + C

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What is the integral of (sqrt(u-1))/u ?
The integral of (sqrt(u-1))/u is 2(-arctan(sqrt(u-1))+sqrt(u-1))+C