What is the integral of (sqrt(x-9))/x ?

The integral of (sqrt(x-9))/x is 6(-arctan(1/3 sqrt(x-9))+1/3 sqrt(x-9))+C

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integral of (sqrt(x-9))/x

\int \frac{x - 9}{x} d x

6 (- arctan (\frac{1}{3} x - 9) + \frac{1}{3} x - 9) + C

Solution steps

\int \frac{x - 9}{x} d x

Apply u-substitution

: \int \frac{2 u ^{2}}{u ^{2} + 9} d u

= \int \frac{2 u ^{2}}{u ^{2} + 9} 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} + 9} d u

\frac{u ^{2}}{u ^{2} + 9} = - \frac{9}{u ^{2} + 9} + 1

= 2 \cdot \int - \frac{9}{u ^{2} + 9} + 1 d u

Apply Integral Substitution

= 2 \cdot \int 3 (- \frac{1}{v ^{2} + 1} + 1) d v

Take the constant out:

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

= 2 \cdot 3 \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 \cdot 3 (- \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 \cdot 3 (- arctan (v) + v)

Substitute back

= 2 \cdot 3 (- arctan (\frac{x - 9}{3}) + \frac{x - 9}{3})

Simplify

2 \cdot 3 (- arctan (\frac{x - 9}{3}) + \frac{x - 9}{3}) : 6 (- arctan (\frac{1}{3} x - 9) + \frac{1}{3} x - 9)

= 6 (- arctan (\frac{1}{3} x - 9) + \frac{1}{3} x - 9)

Add a constant to the solution

= 6 (- arctan (\frac{1}{3} x - 9) + \frac{1}{3} x - 9) + C

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\frac{5 s}{s ^{2} + 2 s + 5}

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What is the integral of (sqrt(x-9))/x ?
The integral of (sqrt(x-9))/x is 6(-arctan(1/3 sqrt(x-9))+1/3 sqrt(x-9))+C