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

The integral of (-x^2)/(x^2+x-2) is -x-1/3 ln|x-1|+4/3 ln|x+2|+C

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

\frac{x ^{2}}{x ^{2} + x - 2} : 1 + \frac{1}{3 ( x - 1 )} - \frac{4}{3 ( x + 2 )}

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

\int 1 d x = x

\int \frac{1}{3 ( x - 1 )} d x = \frac{1}{3} ln ∣ x - 1 ∣

\int \frac{4}{3 ( x + 2 )} d x = \frac{4}{3} ln ∣ x + 2 ∣

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

Simplify

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

Add a constant to the solution

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

What is the integral of (-x^2)/(x^2+x-2) ?
The integral of (-x^2)/(x^2+x-2) is -x-1/3 ln|x-1|+4/3 ln|x+2|+C