What is the integral of 2/(sqrt(x+2)+\sqrt{x)} ?

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

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integral of 2/(sqrt(x+2)+\sqrt{x)}

\int \frac{2}{x + 2 + x} d x

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

Solution steps

\int \frac{2}{x + 2 + x} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{x + 2 + x} d x

Multiply by the conjugate of

x + 2 + x : \frac{2}{x + 2 - x}

= 2 \cdot \int \frac{1}{\frac{2}{x + 2 - x}} d x

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{\frac{1}{x + 2 - x}} d x

Simplify

= 2 \cdot \frac{1}{2} \cdot \int x + 2 - 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

= 2 \cdot \frac{1}{2} (\int x + 2 d x - \int x d x)

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

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

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

Simplify

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

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

Add a constant to the solution

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

e^{9}

\int 3 x + 2 y d y

y^{^{''}} - 8 y^{^{'}} = 64 t, y (0) = 8, y^{^{'}} (0) = 0

\int \frac{x ^{3}}{2 x + 1} d x

\frac{d}{d x} (ln (a x^{2} + b))

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