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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

\frac{x ^{2}}{x + 1} - ln ∣ x + 1 ∣ - \frac{1}{x + 1} - 2 (ln ∣ x ∣ - ln ∣ x + 1 ∣) : \frac{x ^{2}}{x + 1} + ln ∣ x + 1 ∣ - \frac{1}{x + 1} - 2 ln ∣ x ∣

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

Add a constant to the solution

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

\int x^{8} e^{x} d x

(x + 2) \frac{d y}{d x} + y = ln (x), y (1) = 20

x, x^{2}, 0, 1

\frac{\partial}{\partial y} (x^{2} y + y^{2} z)

\frac{x - 14}{x - 14}

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