What is the integral of (1-x)/(x^2-4) ?

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

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integral of (1-x)/(x^2-4)

\int \frac{1 - x}{x ^{2} - 4} d x

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

Solution steps

\int \frac{1 - x}{x ^{2} - 4} d x

Expand

\frac{1 - x}{x ^{2} - 4} : \frac{1}{x ^{2} - 4} - \frac{x}{x ^{2} - 4}

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

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

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

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

Add a constant to the solution

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

f (x) = x^{2} - \frac{4}{x} - 10, at x = 8

6 x^{2}

(2 x + 1)^{2} (0.1)

x^{2} - 2 x, - x^{2} + 2 x

\int (e^{8 x} + 4 e^{- x}) d x

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