What is the integral of 8/(x^4+4) ?

The integral of 8/(x^4+4) is 8(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 8/(x^4+4)

\int \frac{8}{x ^{4} + 4} d x

8 (\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{8}{x ^{4} + 4} d x

Take the constant out:

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

= 8 \cdot \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 )}

= 8 \cdot \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

= 8 (\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))

= 8 (\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

= 8 (\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 8/(x^4+4) ?
The integral of 8/(x^4+4) is 8(1/16 (ln|x^2+2x+2|+2arctan(x+1))+1/16 (-ln|x^2-2x+2|+2arctan(x-1)))+C