What is the integral of 1/((x^2+1)(x^2+3)) ?

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

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integral of 1/((x^2+1)(x^2+3))

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

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