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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int in d x

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