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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u ( 1 - u )} d u

= \int \frac{1}{2 u ( 1 - u )} d u

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}{u ( 1 - u )} d u

Take the partial fraction of

\frac{1}{u ( 1 - u )} : \frac{1}{u} - \frac{1}{u - 1}

= \frac{1}{2} \cdot \int \frac{1}{u} - \frac{1}{u - 1} d u

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}{u} d u - \int \frac{1}{u - 1} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

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

Substitute back

u = x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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