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

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

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

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

Solution steps

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

Apply u-substitution

: \int - 1 + u d u

= \int - 1 + u d u

Take the constant out:

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

= - \int 1 + u d u

Apply u-substitution

: \int v d v

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

Join

1 + \frac{1}{x} : \frac{x + 1}{x}

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

Add a constant to the solution

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

\int (4 x - 2 x) d x

\int 8 + x^{2} d x

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

\int cos (n w x) d x

\int \frac{6 x}{x ^{2} + 3} d x

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