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integral of (x-4)/(x^2+x-2)

Solution

\int \frac{x - 4}{x ^{2} + x - 2} d x

Solution

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

Solution steps

\int \frac{x - 4}{x ^{2} + x - 2} d x

Expand

\frac{x - 4}{x ^{2} + x - 2} : \frac{x}{x ^{2} + x - 2} - \frac{4}{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}{x ^{2} + x - 2} d x - \int \frac{4}{x ^{2} + x - 2} d x

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

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

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

Simplify

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

Add a constant to the solution

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

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\int \frac{1}{y ^{2} + 4 y + 8} d y

integral of sqrt((x^3-4)/(x^{11))}

\int \frac{x ^{3} - 4}{x ^{11}} d x

integral of sin^2(z)

\int sin^{2} (z) d z

Frequently Asked Questions (FAQ)

What is the integral of (x-4)/(x^2+x-2) ?
The integral of (x-4)/(x^2+x-2) is