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

The integral of (x+1)/((x-1)(x^2+x+1)) is 2/3 ln|x-1|-1/3 ln|x^2+x+1|+C

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

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

\frac{2}{3} ln ∣ x - 1 ∣ - \frac{1}{3} ln x^{2} + x + 1 + C

Solution steps

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

Take the partial fraction of

\frac{x + 1}{( x - 1 ) ( x ^{2} + x + 1 )} : \frac{2}{3 ( x - 1 )} + \frac{- 2 x - 1}{3 ( x ^{2} + x + 1 )}

= \int \frac{2}{3 ( x - 1 )} + \frac{- 2 x - 1}{3 ( x ^{2} + 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{2}{3 ( x - 1 )} d x + \int \frac{- 2 x - 1}{3 ( x ^{2} + x + 1 )} d x

\int \frac{2}{3 ( x - 1 )} d x = \frac{2}{3} ln ∣ x - 1 ∣

\int \frac{- 2 x - 1}{3 ( x ^{2} + x + 1 )} d x = - \frac{1}{3} ln x^{2} + x + 1

= \frac{2}{3} ln ∣ x - 1 ∣ - \frac{1}{3} ln x^{2} + x + 1

Add a constant to the solution

= \frac{2}{3} ln ∣ x - 1 ∣ - \frac{1}{3} ln x^{2} + x + 1 + C

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What is the integral of (x+1)/((x-1)(x^2+x+1)) ?
The integral of (x+1)/((x-1)(x^2+x+1)) is 2/3 ln|x-1|-1/3 ln|x^2+x+1|+C