What is the integral of 6/(sqrt(x+6)+\sqrt{x)} ?

The integral of 6/(sqrt(x+6)+\sqrt{x)} is 2/3 (x+6)^{3/2}-2/3 x^{3/2}+C

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integral of 6/(sqrt(x+6)+\sqrt{x)}

\int \frac{6}{x + 6 + x} d x

\frac{2}{3} (x + 6)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}} + C

Solution steps

\int \frac{6}{x + 6 + x} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{x + 6 + x} d x

Multiply by the conjugate of

x + 6 + x : \frac{6}{x + 6 - x}

= 6 \cdot \int \frac{1}{\frac{6}{x + 6 - x}} d x

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int \frac{1}{\frac{1}{x + 6 - x}} d x

Simplify

= 6 \cdot \frac{1}{6} \cdot \int x + 6 - 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

= 6 \cdot \frac{1}{6} (\int x + 6 d x - \int x d x)

\int x + 6 d x = \frac{2}{3} (x + 6)^{\frac{3}{2}}

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

= 6 \cdot \frac{1}{6} (\frac{2}{3} (x + 6)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}})

Simplify

6 \cdot \frac{1}{6} (\frac{2}{3} (x + 6)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}}) : \frac{2}{3} (x + 6)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}}

= \frac{2}{3} (x + 6)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (x + 6)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}} + C

\int g (x) d x

n = 1 \sum \infty arctan (12 n)

\int x (x + 6)^{\frac{1}{2}} d x

\int \frac{1}{5 x + 2} d x

\int_{1}^{2} x e^{3 x^{2}} d x

What is the integral of 6/(sqrt(x+6)+\sqrt{x)} ?
The integral of 6/(sqrt(x+6)+\sqrt{x)} is 2/3 (x+6)^{3/2}-2/3 x^{3/2}+C