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

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

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integral of (1-x^2)/(2x)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

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

= \frac{1}{2} \cdot \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

= \frac{1}{2} (\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}

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

Add a constant to the solution

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

f (t) = e^{3 t}

f (x) = 8 tan (x), (\frac{π}{4}, 8)

\int \frac{e ^{x + 4}}{x + 4} d x

2 y ln (x) ln (y) d x + d y = 0

(4 - 2 x^{2})^{3}

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