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

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

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integral of (x^2+1)/(x^3+x^2-x-1)

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

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

Solution steps

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

Expand

\frac{x ^{2} + 1}{x ^{3} + x ^{2} - x - 1} : \frac{x ^{2}}{x ^{3} + x ^{2} - x - 1} + \frac{1}{x ^{3} + x ^{2} - x - 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 ^{2}}{x ^{3} + x ^{2} - x - 1} d x + \int \frac{1}{x ^{3} + x ^{2} - x - 1} d x

\int \frac{x ^{2}}{x ^{3} + x ^{2} - x - 1} d x = \frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣

\int \frac{1}{x ^{3} + x ^{2} - x - 1} d x = - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣

= \frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣

Simplify

\frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣ : \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{x + 1}

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

Add a constant to the solution

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

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