What is the integral of ((u-1)^{0.5})/u ?

The integral of ((u-1)^{0.5})/u is 2(-arctan(sqrt(u-1))+sqrt(u-1))+C

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integral of ((u-1)^{0.5})/u

integral

\frac{( u - 1 ) ^{0.5}}{u}

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

Solution steps

\int \frac{( u - 1 ) ^{0.5}}{u} d u

Apply u-substitution

: \int \frac{v ^{2}}{0.5 ( v ^{2} + 1 )} d v

= \int \frac{v ^{2}}{0.5 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{0.5} \cdot \int \frac{v ^{2}}{v ^{2} + 1} d v

Simplify

= 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)^{0.5}

= 2 (- arctan ((u - 1)^{0.5}) + (u - 1)^{0.5})

Apply radical rule:

a^{\frac{1}{n}} = n a

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

Add a constant to the solution

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

\frac{( y + 1 ) ^{2}}{y}

(1 - y)^{4}

(2 x + 1)^{- 3}

\frac{1}{( 9 x ^{2} + 4 )}

y = \frac{1}{x ^{\frac{1}{2}}}

What is the integral of ((u-1)^{0.5})/u ?
The integral of ((u-1)^{0.5})/u is 2(-arctan(sqrt(u-1))+sqrt(u-1))+C