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

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

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

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

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

Solution steps

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

Expand

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

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

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

\int \frac{1}{x ^{2} + x - 2} d x = - \frac{1}{3} (ln \frac{4}{3} + \frac{2}{3} x - ln - \frac{2}{3} + \frac{2}{3} x)

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

Add a constant to the solution

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

y^{^{'}} (t) = (y + 1) (y - 3), y (0) = 1

t^{3} \frac{d y}{d t} + 3 t^{2} y = 7 cos (t), y (π) = 0

x \to \infty lim (- \frac{x}{s})

18 (2 e^{x^{2}} x^{2} + e^{x^{2}})

\frac{x - 3}{x + 2}

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