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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

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