What is the integral of (x^2-4)/(x^2-16) ?

The integral of (x^2-4)/(x^2-16) is -3/2 ln| x/4+1|+3/2 ln| x/4-1|+x+C

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integral of (x^2-4)/(x^2-16)

\int \frac{x ^{2} - 4}{x ^{2} - 16} d x

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

Solution steps

\int \frac{x ^{2} - 4}{x ^{2} - 16} d x

Apply Integral Substitution

= \int \frac{2 ( u ^{2} - 1 )}{u ^{2} - 4} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2} - 1}{u ^{2} - 4} d u

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d u = u

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

Substitute back

u = \frac{x}{2}

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (x^2-4)/(x^2-16) ?
The integral of (x^2-4)/(x^2-16) is -3/2 ln| x/4+1|+3/2 ln| x/4-1|+x+C