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

The integral of (x-x^{-3})/(x^2+x^{-2)} is 1/2 ln|x^4+1|-ln(x)+C

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integral of (x-x^{-3})/(x^2+x^{-2)}

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

\frac{1}{2} ln x^{4} + 1 - ln (x) + C

Solution steps

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

Expand

\frac{x - x ^{- 3}}{x ^{2} + x ^{- 2}} : \frac{x}{x ^{2} + x ^{- 2}} - \frac{x ^{- 3}}{x ^{2} + x ^{- 2}}

Apply the Sum Rule:

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

= \int \frac{x}{x ^{2} + x ^{- 2}} d x - \int \frac{x ^{- 3}}{x ^{2} + x ^{- 2}} d x

\int \frac{x}{x ^{2} + x ^{- 2}} d x = \frac{1}{4} ln x^{4} + 1

\int \frac{x ^{- 3}}{x ^{2} + x ^{- 2}} d x = \frac{1}{4} (4 ln (x) - ln 1 + x^{4})

= \frac{1}{4} ln x^{4} + 1 - \frac{1}{4} (4 ln (x) - ln 1 + x^{4})

Simplify

\frac{1}{4} ln x^{4} + 1 - \frac{1}{4} (4 ln (x) - ln 1 + x^{4}) : \frac{1}{2} ln x^{4} + 1 - ln (x)

= \frac{1}{2} ln x^{4} + 1 - ln (x)

Add a constant to the solution

= \frac{1}{2} ln x^{4} + 1 - ln (x) + C

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What is the integral of (x-x^{-3})/(x^2+x^{-2)} ?
The integral of (x-x^{-3})/(x^2+x^{-2)} is 1/2 ln|x^4+1|-ln(x)+C