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

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

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

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

Solution steps

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

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

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

= \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = z + z^{- 1}

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

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

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

Add a constant to the solution

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

\int t sin (5 t) d t

\int \frac{( 4 x - 1 )}{( 2 x + 1 )} d x

\int \frac{x ^{2}}{( x + 5 ) ^{3}} d x

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

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