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

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

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integral of (1-x^2)/x

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

ln ∣ x ∣ - \frac{x ^{2}}{2} + C

Solution steps

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

Expand

\frac{1 - x ^{2}}{x} : \frac{1}{x} - x

= \int \frac{1}{x} - x 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 \frac{1}{x} d x - \int x d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - \frac{x ^{2}}{2}

Add a constant to the solution

= ln ∣ x ∣ - \frac{x ^{2}}{2} + C

f (x) = 7^{x}

\frac{2}{3} t^{3} \cdot e^{- 8 t}

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

3 y^{3} e^{x} d x + e^{- x} d y = 3 y^{3} d x

\frac{d}{d x} (ln (1 - \frac{1}{x}))

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