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

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

Solution steps

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

Take the partial fraction of

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

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

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

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

= x + ln ∣ x ∣ - 3 ln ∣ x + 1 ∣

Add a constant to the solution

= x + ln ∣ x ∣ - 3 ln ∣ x + 1 ∣ + C

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