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

2 x^{2}, 8 x^{2}, 5 - 3 x

\frac{\partial}{\partial x} (4 x^{2} y + 7 x y^{3})

f (x) = (3 x - 2) (4 x^{3} - x^{2} + 1)

\frac{d}{d x} (6 x e^{x^{2}})

x \to 2 lim (x^{2} - 6 x + 3)

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