What is the integral of (4x)/(sqrt(x^2-1)) ?

The integral of (4x)/(sqrt(x^2-1)) is 4sqrt(x^2-1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x)/(sqrt(x^2-1))

\int \frac{4 x}{x ^{2} - 1} d x

4 x^{2} - 1 + C

Solution steps

\int \frac{4 x}{x ^{2} - 1} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{x ^{2} - 1} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Take the constant out:

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= 4 \cdot \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{2} - 1

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

Simplify

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

= 4 x^{2} - 1

Add a constant to the solution

= 4 x^{2} - 1 + C

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

\int \frac{x}{x ^{4} - 3 x ^{2} + 2} d x

(sin (\frac{1}{x}))^{^{'}}

f (x) = ln (19 x^{3} + sin (3 x^{2}))

\frac{d}{d X} (X^{3} - 3 X)

What is the integral of (4x)/(sqrt(x^2-1)) ?
The integral of (4x)/(sqrt(x^2-1)) is 4sqrt(x^2-1)+C