What is the integral of 2/(sqrt(x)+4x\sqrt{x)} ?

The integral of 2/(sqrt(x)+4x\sqrt{x)} is 2arctan(2sqrt(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 2/(sqrt(x)+4x\sqrt{x)}

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

2 arctan (2 x) + C

Solution steps

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

\frac{1}{x + 4 x x} = \frac{x - 4 x x}{( x ) ^{2} - ( 4 x x ) ^{2}}

= 2 \cdot \int \frac{x - 4 x x}{( x ) ^{2} - ( 4 x x ) ^{2}} d x

Simplify

\frac{x - 4 x x}{( x ) ^{2} - ( 4 x x ) ^{2}} : \frac{1}{x ( 4 x + 1 )}

= 2 \cdot \int \frac{1}{x ( 4 x + 1 )} d x

Apply u-substitution

= 2 \cdot \int \frac{2}{4 u ^{2} + 1} d u

Take the constant out:

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

= 2 \cdot 2 \cdot \int \frac{1}{4 u ^{2} + 1} d u

Apply Integral Substitution

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

Substitute back

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

Simplify

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

= 2 arctan (2 x)

Add a constant to the solution

= 2 arctan (2 x) + C

What is the integral of 2/(sqrt(x)+4x\sqrt{x)} ?
The integral of 2/(sqrt(x)+4x\sqrt{x)} is 2arctan(2sqrt(x))+C