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

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

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

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

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

Solution steps

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

Expand

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

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

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

= \frac{1}{2} arctan (\frac{x}{2}) + 2 (- \frac{1}{4 x} - \frac{1}{8} arctan (\frac{x}{2}))

Simplify

\frac{1}{2} arctan (\frac{x}{2}) + 2 (- \frac{1}{4 x} - \frac{1}{8} arctan (\frac{x}{2})) : \frac{1}{4} arctan (\frac{x}{2}) - \frac{1}{2 x}

= \frac{1}{4} arctan (\frac{x}{2}) - \frac{1}{2 x}

Add a constant to the solution

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

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