What is the integral of (10)/(sqrt(x+10)+\sqrt{x)} ?

The integral of (10)/(sqrt(x+10)+\sqrt{x)} is 2/3 (x+10)^{3/2}-2/3 x^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (10)/(sqrt(x+10)+\sqrt{x)}

\int \frac{10}{x + 10 + x} d x

\frac{2}{3} (x + 10)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}} + C

Solution steps

\int \frac{10}{x + 10 + x} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{x + 10 + x} d x

Multiply by the conjugate of

x + 10 + x : \frac{10}{x + 10 - x}

= 10 \cdot \int \frac{1}{\frac{10}{x + 10 - x}} d x

Take the constant out:

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

= 10 \cdot \frac{1}{10} \cdot \int \frac{1}{\frac{1}{x + 10 - x}} d x

Simplify

= 10 \cdot \frac{1}{10} \cdot \int x + 10 - 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

= 10 \cdot \frac{1}{10} (\int x + 10 d x - \int x d x)

\int x + 10 d x = \frac{2}{3} (x + 10)^{\frac{3}{2}}

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

= 10 \cdot \frac{1}{10} (\frac{2}{3} (x + 10)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}})

Simplify

10 \cdot \frac{1}{10} (\frac{2}{3} (x + 10)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}}) : \frac{2}{3} (x + 10)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}}

= \frac{2}{3} (x + 10)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (x + 10)^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}} + C

What is the integral of (10)/(sqrt(x+10)+\sqrt{x)} ?
The integral of (10)/(sqrt(x+10)+\sqrt{x)} is 2/3 (x+10)^{3/2}-2/3 x^{3/2}+C