What is the integral of 8/(sqrt(x+8)+\sqrt{x)} ?

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

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integral of 8/(sqrt(x+8)+\sqrt{x)}

\int \frac{8}{x + 8 + x} d x

\frac{2}{3} (x + 8)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}} + C

Solution steps

\int \frac{8}{x + 8 + x} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{x + 8 + x} d x

Multiply by the conjugate of

x + 8 + x : \frac{8}{x + 8 - x}

= 8 \cdot \int \frac{1}{\frac{8}{x + 8 - x}} d x

Take the constant out:

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

= 8 \cdot \frac{1}{8} \cdot \int \frac{1}{\frac{1}{x + 8 - x}} d x

Simplify

= 8 \cdot \frac{1}{8} \cdot \int x + 8 - 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

= 8 \cdot \frac{1}{8} (\int x + 8 d x - \int x d x)

\int x + 8 d x = \frac{2}{3} (x + 8)^{\frac{3}{2}}

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

= 8 \cdot \frac{1}{8} (\frac{2}{3} (x + 8)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}})

Simplify

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

= \frac{2}{3} (x + 8)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (x + 8)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}} + C

f (z) = e^{\frac{z}{( z - 1 )}}

\frac{d y}{d x} = \frac{x}{2 y ^{2}}

x \to 8 + lim (\frac{7 - x}{8 - x})

\frac{d y}{d x} = \frac{cos ( 11 x )}{e ^{11 y}}

x \to 5 lim (\frac{3}{( x - 5 ) ^{2}})

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