What is the integral of (x+x^{-1}) ?

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

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integral of (x+x^{-1})

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

\frac{x ^{2}}{2} + ln ∣ x ∣ + C

Solution steps

\int x + x^{- 1} 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 d x + \int x^{- 1} d x

\int x d x = \frac{x ^{2}}{2}

\int x^{- 1} d x = ln ∣ x ∣

= \frac{x ^{2}}{2} + ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{2} + ln ∣ x ∣ + C

\int \frac{1}{x ^{2} 13 - x ^{2}} d x

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

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

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

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