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

The integral of 1/((x^4+4)) is 1/16 (ln|x^2+2x+2|+2arctan(x+1))+1/16 (-ln|x^2-2x+2|+2arctan(x-1))+C

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

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

\frac{1}{16} (ln x^{2} + 2 x + 2 + 2 arctan (x + 1)) + \frac{1}{16} (- ln x^{2} - 2 x + 2 + 2 arctan (x - 1)) + C

Solution steps

\int \frac{1}{x ^{4} + 4} d x

Take the partial fraction of

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

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

\int \frac{x + 2}{8 ( x ^{2} + 2 x + 2 )} d x = \frac{1}{16} (ln x^{2} + 2 x + 2 + 2 arctan (x + 1))

\int \frac{- x + 2}{8 ( x ^{2} - 2 x + 2 )} d x = \frac{1}{16} (- ln x^{2} - 2 x + 2 + 2 arctan (x - 1))

= \frac{1}{16} (ln x^{2} + 2 x + 2 + 2 arctan (x + 1)) + \frac{1}{16} (- ln x^{2} - 2 x + 2 + 2 arctan (x - 1))

Add a constant to the solution

= \frac{1}{16} (ln x^{2} + 2 x + 2 + 2 arctan (x + 1)) + \frac{1}{16} (- ln x^{2} - 2 x + 2 + 2 arctan (x - 1)) + C

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What is the integral of 1/((x^4+4)) ?
The integral of 1/((x^4+4)) is 1/16 (ln|x^2+2x+2|+2arctan(x+1))+1/16 (-ln|x^2-2x+2|+2arctan(x-1))+C