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

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

Popular >

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

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

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

Solution steps

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

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

x \to \infty lim (\frac{3 e ^{x}}{9 + 2 e ^{5 x}})

y^{^{'}} + 3 y = 5

x \to \infty lim (\frac{x ^{3} + 2 x + 4}{e ^{1 x} + 1})

y = 4 x^{2} - 6 x, at x = 2

\frac{d y}{d x} = \frac{y ^{2} + x ^{2}}{y x}

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