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

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

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

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

x + 3 ln ∣ x - 1 ∣ - ln ∣ x ∣ - 1 + C

Solution steps

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

Expand

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

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

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

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

= x - 1 + ln ∣ x - 1 ∣ + ln ∣ x - 1 ∣ - ln ∣ x ∣ + ln ∣ x - 1 ∣

Simplify

= x + 3 ln ∣ x - 1 ∣ - ln ∣ x ∣ - 1

Add a constant to the solution

= x + 3 ln ∣ x - 1 ∣ - ln ∣ x ∣ - 1 + C

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 7} e^{5 t}

\int \frac{6 x ^{4} + 23 x ^{2} + 4 x + 15}{x ^{2} + 3} d x

\frac{d}{d x} (x^{2} sin^{7} (x) + x cos^{- 2} (x))

4 y^{^{''}} + 2 y^{^{'}} + 2 y = 0

y = cos (x), at x = \frac{π}{2}

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