What is the integral of (x^b-x^{-b})^2 ?

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

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integral of (x^b-x^{-b})^2

\int (x^{b} - x^{- b})^{2} d x

\frac{x ^{2 b + 1}}{2 b + 1} - 2 x + \frac{x ^{- 2 b + 1}}{- 2 b + 1} + C

Solution steps

\int (x^{b} - x^{- b})^{2} d x

Expand

(x^{b} - x^{- b})^{2} : x^{2 b} - 2 + x^{- 2 b}

= \int x^{2 b} - 2 + x^{- 2 b} d x

Apply the Sum Rule:

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

= \int x^{2 b} d x - \int 2 d x + \int x^{- 2 b} d x

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

\int 2 d x = 2 x

\int x^{- 2 b} d x = \frac{x ^{- 2 b + 1}}{- 2 b + 1}

= \frac{x ^{2 b + 1}}{2 b + 1} - 2 x + \frac{x ^{- 2 b + 1}}{- 2 b + 1}

Add a constant to the solution

= \frac{x ^{2 b + 1}}{2 b + 1} - 2 x + \frac{x ^{- 2 b + 1}}{- 2 b + 1} + C

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