What is the integral of (5-x)/(x^2+x-2) ?

The integral of (5-x)/(x^2+x-2) is 4/3 ln|x-1|-7/3 ln|x+2|+C

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integral of (5-x)/(x^2+x-2)

\int \frac{5 - x}{x ^{2} + x - 2} d x

\frac{4}{3} ln ∣ x - 1 ∣ - \frac{7}{3} ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{5 - x}{x ^{2} + x - 2} d x

Take the partial fraction of

\frac{5 - x}{x ^{2} + x - 2} : \frac{4}{3 ( x - 1 )} - \frac{7}{3 ( x + 2 )}

= \int \frac{4}{3 ( x - 1 )} - \frac{7}{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 \frac{4}{3 ( x - 1 )} d x - \int \frac{7}{3 ( x + 2 )} d x

\int \frac{4}{3 ( x - 1 )} d x = \frac{4}{3} ln ∣ x - 1 ∣

\int \frac{7}{3 ( x + 2 )} d x = \frac{7}{3} ln ∣ x + 2 ∣

= \frac{4}{3} ln ∣ x - 1 ∣ - \frac{7}{3} ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{4}{3} ln ∣ x - 1 ∣ - \frac{7}{3} ln ∣ x + 2 ∣ + C

\frac{d}{d x} (e^{- 4 x} sin (4 x))

\int \frac{tan ^{2} ( θ )}{sec ( θ )} d θ

- y + \frac{d y}{d x} ln (x) = 0

\frac{d}{d x} (arcsin (x^{2} + 1))

y = \frac{x}{36 + x ^{2}}, (0, 0)

What is the integral of (5-x)/(x^2+x-2) ?
The integral of (5-x)/(x^2+x-2) is 4/3 ln|x-1|-7/3 ln|x+2|+C