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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\int \frac{( z - 1 ) ^{2}}{z} d z

\frac{2}{3} z^{\frac{3}{2}} - 2 z + 2 z + C

Solution steps

\int \frac{( z - 1 ) ^{2}}{z} d z

Expand

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

= \int z - 2 + \frac{1}{z} d z

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int z d z - \int 2 d z + \int \frac{1}{z} d z

\int z d z = \frac{2}{3} z^{\frac{3}{2}}

\int 2 d z = 2 z

\int \frac{1}{z} d z = 2 z

= \frac{2}{3} z^{\frac{3}{2}} - 2 z + 2 z

Add a constant to the solution

= \frac{2}{3} z^{\frac{3}{2}} - 2 z + 2 z + C

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