What is the integral of 9/(sqrt(x+9)+\sqrt{x)} ?

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

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integral of 9/(sqrt(x+9)+\sqrt{x)}

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

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

Solution steps

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

Take the constant out:

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

= 9 \cdot \int \frac{1}{x + 9 + x} d x

Multiply by the conjugate of

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

= 9 \cdot \int \frac{1}{\frac{9}{x + 9 - x}} d x

Take the constant out:

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

= 9 \cdot \frac{1}{9} \cdot \int \frac{1}{\frac{1}{x + 9 - x}} d x

Simplify

= 9 \cdot \frac{1}{9} \cdot \int x + 9 - x 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

= 9 \cdot \frac{1}{9} (\int x + 9 d x - \int x d x)

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

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

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

Simplify

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

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

Add a constant to the solution

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

f (x) = x^{2} (1 - 8 x)

ln (2 x + 1)

x^{2} - 5 x, x + 7

f^{^{''}} (x) = \frac{x ^{2} - 2 x + 1}{x ^{2}}

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

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