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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of sqrt(4/x-1)

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

x 4 - x - 4 arctan (\frac{4 - x}{x}) + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply Integration By Parts

= - 4 (- \frac{u - 1}{u} - \int - \frac{1}{2 u u - 1} d u)

\int - \frac{1}{2 u u - 1} d u = - arctan (u - 1)

= - 4 (- \frac{u - 1}{u} - (- arctan (u - 1)))

Substitute back

u = \frac{4}{x}

= - 4 - \frac{\frac{4}{x} - 1}{\frac{4}{x}} - (- arctan (\frac{4}{x} - 1))

Simplify

- 4 - \frac{\frac{4}{x} - 1}{\frac{4}{x}} - (- arctan (\frac{4}{x} - 1)) : x 4 - x - 4 arctan (\frac{4 - x}{x})

= x 4 - x - 4 arctan (\frac{4 - x}{x})

Add a constant to the solution

= x 4 - x - 4 arctan (\frac{4 - x}{x}) + C

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

\int x (x + 1)^{\frac{2}{3}} d x

\int 3 7 x d x

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

\int \frac{9 x ( x ^{2} - 10 ) ^{8}}{5} d x

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