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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1

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

Simplify

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