What is the integral of (sqrt(x))/(sqrt(x)-3) ?

The integral of (sqrt(x))/(sqrt(x)-3) is x+6sqrt(x)+18ln|sqrt(x)-3|-27+C

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integral of (sqrt(x))/(sqrt(x)-3)

\int \frac{x}{x - 3} d x

x + 6 x + 18 ln x - 3 - 27 + C

Solution steps

\int \frac{x}{x - 3} d x

Apply u-substitution

: \int \frac{2 ( u ^{2} + 6 u + 9 )}{u} d u

= \int \frac{2 ( u ^{2} + 6 u + 9 )}{u} 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} + 6 u + 9}{u} d u

Expand

\frac{u ^{2} + 6 u + 9}{u} : u + 6 + \frac{9}{u}

= 2 \cdot \int u + 6 + \frac{9}{u} 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

= 2 (\int u d u + \int 6 d u + \int \frac{9}{u} d u)

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

\int 6 d u = 6 u

\int \frac{9}{u} d u = 9 ln ∣ u ∣

= 2 (\frac{u ^{2}}{2} + 6 u + 9 ln ∣ u ∣)

Substitute back

u = x - 3

= 2 (\frac{( x - 3 ) ^{2}}{2} + 6 (x - 3) + 9 ln x - 3)

Simplify

2 (\frac{( x - 3 ) ^{2}}{2} + 6 (x - 3) + 9 ln x - 3) : x + 6 x + 18 ln x - 3 - 27

= x + 6 x + 18 ln x - 3 - 27

Add a constant to the solution

= x + 6 x + 18 ln x - 3 - 27 + C

\int \frac{1}{2 x ^{4}} d x

\int (8 x + 5) 4 x^{2} + 5 x d x

- \frac{1}{x ^{2}}, 2

y^{^{'}} = 0

What is the integral of (sqrt(x))/(sqrt(x)-3) ?
The integral of (sqrt(x))/(sqrt(x)-3) is x+6sqrt(x)+18ln|sqrt(x)-3|-27+C