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

The integral of x/(sqrt(x-9)) is 2(1/3 (x-9)^{3/2}+9sqrt(x-9))+C

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

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

2 (\frac{1}{3} (x - 9)^{\frac{3}{2}} + 9 x - 9) + C

Solution steps

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

Apply u-substitution

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

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

\int u^{2} d u = \frac{u ^{3}}{3}

\int 9 d u = 9 u

= 2 (\frac{u ^{3}}{3} + 9 u)

Substitute back

u = x - 9

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

Simplify

2 (\frac{( x - 9 ) ^{3}}{3} + 9 x - 9) : 2 (\frac{1}{3} (x - 9)^{\frac{3}{2}} + 9 x - 9)

= 2 (\frac{1}{3} (x - 9)^{\frac{3}{2}} + 9 x - 9)

Add a constant to the solution

= 2 (\frac{1}{3} (x - 9)^{\frac{3}{2}} + 9 x - 9) + C

4 x^{2} + 2 x - 1, at x = 0

x y^{2} y^{^{'}} = x^{3} + 2 y^{3}

\frac{30}{s ( s + 5 ) ( s + 4 )}

\frac{d}{d x} (\frac{2 x}{3 x ^{2} - 3})

\frac{d y}{d x} = (y e^{x})

What is the integral of x/(sqrt(x-9)) ?
The integral of x/(sqrt(x-9)) is 2(1/3 (x-9)^{3/2}+9sqrt(x-9))+C