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

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

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

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

- arctan (x - 1) + x - 1 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

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

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

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

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

\int 1 d u = u

= \frac{1}{2} \cdot 2 (- arctan (u) + u)

Substitute back

u = x - 1

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

Simplify

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

= - arctan (x - 1) + x - 1

Add a constant to the solution

= - arctan (x - 1) + x - 1 + C

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