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

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

Popular >

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) = 5 x^{2} - 8 x + 3, at x = 1

x \to 0 lim (tan (\frac{x}{2}))

x arcsin (x)

n = 1 \sum \infty n!^{2} (5^{n})

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

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