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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

= 4 \cdot \int \frac{1}{4 x ^{2}} - \frac{1}{4 ( 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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

x^{3}, 3

\int x^{10} e^{x^{11}} d x

\frac{d}{d x} (7 sin^{2} (x))

f (x) = 1 - 3 x, at x = - 1

\frac{d}{d x} (\frac{3 x ^{2} - 1}{x - 2})

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