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

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

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

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

- 2 ln ∣ x + 1 ∣ + ln ∣ x ∣ + x + C

Solution steps

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

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

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

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

\int 1 d x = x

= - 2 ln ∣ x + 1 ∣ + ln ∣ x ∣ + x

Add a constant to the solution

= - 2 ln ∣ x + 1 ∣ + ln ∣ x ∣ + x + C

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