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

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

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

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

22 (- arctan (\frac{1}{2} (x - 2)) + \frac{1}{2} (x - 2)) + C

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

= 2 \cdot \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

= 22 \cdot \int \frac{v ^{2}}{v ^{2} + 1} d v

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

= 22 \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

= 22 (- \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

= 22 (- arctan (v) + v)

Substitute back

= 22 (- arctan (\frac{x - 2}{2}) + \frac{x - 2}{2})

Simplify

22 (- arctan (\frac{x - 2}{2}) + \frac{x - 2}{2}) : 22 (- arctan (\frac{1}{2} (x - 2)) + \frac{1}{2} (x - 2))

= 22 (- arctan (\frac{1}{2} (x - 2)) + \frac{1}{2} (x - 2))

Add a constant to the solution

= 22 (- arctan (\frac{1}{2} (x - 2)) + \frac{1}{2} (x - 2)) + C

18 t

y = - 2 e^{- x^{2}} x

\int \frac{3 x}{1 + x ^{3}} d x

y^{^{'}} + 2 x y = y + 4 x - 2

\frac{d}{d x} (\frac{1}{1 + e ^{x}})

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