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

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

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

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

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

Solution steps

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

Long division

\frac{x ( x ^{2} + 4 )}{x ^{2} - 4} : x + \frac{8 x}{x ^{2} - 4}

= \int x + \frac{8 x}{x ^{2} - 4} 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 x d x + \int \frac{8 x}{x ^{2} - 4} d x

\int x d x = \frac{x ^{2}}{2}

\int \frac{8 x}{x ^{2} - 4} d x = 4 ln x^{2} - 4

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

Add a constant to the solution

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

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