What is the integral of (x^4+x)/(x^3-x^2+x-1) ?

The integral of (x^4+x)/(x^3-x^2+x-1) is (x^2)/2+x+ln|x-1|-1/2 ln|x^2+1|+C

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integral of (x^4+x)/(x^3-x^2+x-1)

\int \frac{x ^{4} + x}{x ^{3} - x ^{2} + x - 1} d x

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

Solution steps

\int \frac{x ^{4} + x}{x ^{3} - x ^{2} + x - 1} d x

Long division

\frac{x ^{4} + x}{x ^{3} - x ^{2} + x - 1} : x + 1 + \frac{x + 1}{x ^{3} - x ^{2} + x - 1}

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

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

\int 1 d x = x

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

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

Add a constant to the solution

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

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What is the integral of (x^4+x)/(x^3-x^2+x-1) ?
The integral of (x^4+x)/(x^3-x^2+x-1) is (x^2)/2+x+ln|x-1|-1/2 ln|x^2+1|+C