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

The integral of (x^4+x^2+4)/(x^2-4) is (x^3)/3+5x-6ln| x/2+1|+6ln| x/2-1|+C

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

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

\frac{x ^{3}}{3} + 5 x - 6 ln \frac{x}{2} + 1 + 6 ln \frac{x}{2} - 1 + C

Solution steps

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

Expand

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

\int \frac{x ^{4}}{x ^{2} - 4} d x = - 8 (- \frac{x ^{3}}{24} - \frac{x}{2} + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1)

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

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

= - 8 (- \frac{x ^{3}}{24} - \frac{x}{2} + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1) - ln \frac{x}{2} + 1 + ln \frac{x}{2} - 1 + x - ln \frac{x}{2} + 1 + ln \frac{x}{2} - 1

Simplify

- 8 (- \frac{x ^{3}}{24} - \frac{x}{2} + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1) - ln \frac{x}{2} + 1 + ln \frac{x}{2} - 1 + x - ln \frac{x}{2} + 1 + ln \frac{x}{2} - 1 : \frac{x ^{3}}{3} + 5 x - 6 ln \frac{x}{2} + 1 + 6 ln \frac{x}{2} - 1

= \frac{x ^{3}}{3} + 5 x - 6 ln \frac{x}{2} + 1 + 6 ln \frac{x}{2} - 1

Add a constant to the solution

= \frac{x ^{3}}{3} + 5 x - 6 ln \frac{x}{2} + 1 + 6 ln \frac{x}{2} - 1 + C

\int \frac{4 x ^{2}}{x ^{3} + 16} d x

y = x^{\frac{5}{3}} - x^{\frac{2}{3}}

x \to - 4 lim (x (x + 3) (x + 2))

\frac{x}{- 6 x + 6}

3 sin (x), 3 cos (x)

What is the integral of (x^4+x^2+4)/(x^2-4) ?
The integral of (x^4+x^2+4)/(x^2-4) is (x^3)/3+5x-6ln| x/2+1|+6ln| x/2-1|+C