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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

Add a constant to the solution

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

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