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

The integral of 1/(x^2)sqrt(2-1/x) is 2/3 ((2x-1)/x)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{2}{3} (\frac{2 x - 1}{x})^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int - 2 - u d u

= \int - 2 - u d u

Take the constant out:

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

= - \int 2 - u d u

Apply u-substitution

: \int - v d v

= - \int - v d v

Take the constant out:

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

= - (- \int v d v)

Apply the Power Rule

: \frac{2}{3} v^{\frac{3}{2}}

= - (- \frac{2}{3} v^{\frac{3}{2}})

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

(19 - (15 - 2 x^{4})^{7})^{3}

(12 x^{2} + y - 1) d x - (6 y - x) d y = 0

x y^{^{'}} + 5 y = 0

f (x) = e^{x y}

x \to 0 lim (\frac{5 ^{x} - 1}{7 x})

What is the integral of 1/(x^2)sqrt(2-1/x) ?
The integral of 1/(x^2)sqrt(2-1/x) is 2/3 ((2x-1)/x)^{3/2}+C