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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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