What is the integral of 7/(sqrt(x+7)+\sqrt{x)} ?

The integral of 7/(sqrt(x+7)+\sqrt{x)} is 2/3 (x+7)^{3/2}-2/3 x^{3/2}+C

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integral of 7/(sqrt(x+7)+\sqrt{x)}

\int \frac{7}{x + 7 + x} d x

\frac{2}{3} (x + 7)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}} + C

Solution steps

\int \frac{7}{x + 7 + x} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{x + 7 + x} d x

Multiply by the conjugate of

x + 7 + x : \frac{7}{x + 7 - x}

= 7 \cdot \int \frac{1}{\frac{7}{x + 7 - x}} d x

Take the constant out:

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

= 7 \cdot \frac{1}{7} \cdot \int \frac{1}{\frac{1}{x + 7 - x}} d x

Simplify

= 7 \cdot \frac{1}{7} \cdot \int x + 7 - 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

= 7 \cdot \frac{1}{7} (\int x + 7 d x - \int x d x)

\int x + 7 d x = \frac{2}{3} (x + 7)^{\frac{3}{2}}

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

= 7 \cdot \frac{1}{7} (\frac{2}{3} (x + 7)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}})

Simplify

7 \cdot \frac{1}{7} (\frac{2}{3} (x + 7)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}}) : \frac{2}{3} (x + 7)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}}

= \frac{2}{3} (x + 7)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (x + 7)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}} + C

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

\int \frac{8 x}{x ^{2} + 4} d x

n = 1 \sum \infty \frac{4 n + 2}{5 ^{n}}

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

\frac{d y}{d x} + \frac{1}{x + 1} y = x^{- 2}

What is the integral of 7/(sqrt(x+7)+\sqrt{x)} ?
The integral of 7/(sqrt(x+7)+\sqrt{x)} is 2/3 (x+7)^{3/2}-2/3 x^{3/2}+C