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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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