What is the integral of 1/(sqrt(x)(x+2)) ?

The integral of 1/(sqrt(x)(x+2)) is sqrt(2)arctan(sqrt(x/2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(x)(x+2))

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

2 arctan (\frac{x}{2}) + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{u ^{2} + 2} d u

= \int \frac{2}{u ^{2} + 2} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{u ^{2} + 2} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{2 ( v ^{2} + 1 )} d v

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}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 \cdot \frac{1}{2} arctan (v)

Substitute back

= 2 \cdot \frac{1}{2} arctan (\frac{x}{2})

Simplify

2 \cdot \frac{1}{2} arctan (\frac{x}{2}) : 2 arctan (\frac{x}{2})

= 2 arctan (\frac{x}{2})

Add a constant to the solution

= 2 arctan (\frac{x}{2}) + C

(2 y + 2 t - 7) d t + (4 y + 2 t - 1) d y = 0

\frac{\partial}{\partial u} (tan (5 u v^{5}))

\int \frac{1}{( 16 - x ^{2} ) ^{2}} d x

\frac{d}{d x} (x^{4} sin (2 x))

sin^{2} (e^{s i n^{2} (x)})

What is the integral of 1/(sqrt(x)(x+2)) ?
The integral of 1/(sqrt(x)(x+2)) is sqrt(2)arctan(sqrt(x/2))+C