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

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

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

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

\frac{x ^{4}}{4} + \frac{x ^{3}}{3} + x^{2} + 2 x + 3 ln ∣ x - 1 ∣ + C

Solution steps

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

Long division

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

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

\int x^{3} d x = \frac{x ^{4}}{4}

\int x^{2} d x = \frac{x ^{3}}{3}

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

\int 2 d x = 2 x

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

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

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{x ^{3}}{3} + x^{2} + 2 x + 3 ln ∣ x - 1 ∣ + C

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