What is the integral of (x^2-8x-9)/(sqrt(x)-3) ?

The integral of (x^2-8x-9)/(sqrt(x)-3) is 2/5 x^{5/2}+(3x^2)/2+2/3 x^{3/2}+3x+C

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integral of (x^2-8x-9)/(sqrt(x)-3)

\int \frac{x ^{2} - 8 x - 9}{x - 3} d x

\frac{2}{5} x^{\frac{5}{2}} + \frac{3 x ^{2}}{2} + \frac{2}{3} x^{\frac{3}{2}} + 3 x + C

Solution steps

\int \frac{x ^{2} - 8 x - 9}{x - 3} d x

Multiply by the conjugate of

x - 3 : \frac{x - 9}{x + 3}

= \int \frac{x ^{2} - 8 x - 9}{\frac{x - 9}{x + 3}} d x

Simplify

\frac{x ^{2} - 8 x - 9}{\frac{x - 9}{x + 3}} : (x + 1) (x + 3)

= \int (x + 1) (x + 3) d x

Expand

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

= \int x^{\frac{3}{2}} + 3 x + x + 3 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{\frac{3}{2}} d x + \int 3 x d x + \int x d x + \int 3 d x

\int x^{\frac{3}{2}} d x = \frac{2}{5} x^{\frac{5}{2}}

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

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

\int 3 d x = 3 x

= \frac{2}{5} x^{\frac{5}{2}} + \frac{3 x ^{2}}{2} + \frac{2}{3} x^{\frac{3}{2}} + 3 x

Add a constant to the solution

= \frac{2}{5} x^{\frac{5}{2}} + \frac{3 x ^{2}}{2} + \frac{2}{3} x^{\frac{3}{2}} + 3 x + C

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What is the integral of (x^2-8x-9)/(sqrt(x)-3) ?
The integral of (x^2-8x-9)/(sqrt(x)-3) is 2/5 x^{5/2}+(3x^2)/2+2/3 x^{3/2}+3x+C