What is the integral of 1/((x^2+1)(x^2+4)) ?

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

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integral of 1/((x^2+1)(x^2+4))

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

\frac{1}{3} arctan (x) - \frac{1}{6} arctan (\frac{x}{2}) + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x ^{2} + 1 ) ( x ^{2} + 4 )} : \frac{1}{3 ( x ^{2} + 1 )} - \frac{1}{3 ( x ^{2} + 4 )}

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

\int \frac{1}{3 ( x ^{2} + 1 )} d x = \frac{1}{3} arctan (x)

\int \frac{1}{3 ( x ^{2} + 4 )} d x = \frac{1}{6} arctan (\frac{x}{2})

= \frac{1}{3} arctan (x) - \frac{1}{6} arctan (\frac{x}{2})

Add a constant to the solution

= \frac{1}{3} arctan (x) - \frac{1}{6} arctan (\frac{x}{2}) + C

\int \frac{x}{8 x - 7} d x

\frac{\partial}{\partial z} (6 x^{\frac{y}{z}})

\frac{1}{s + 3} - \frac{1}{s + 4}

\frac{\partial}{\partial s} (\frac{( 2 r + s )}{t})

\frac{d}{d y} (x sin (x y))

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