What is the integral of (32)/(x^4+16) ?

The integral of (32)/(x^4+16) is

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integral of (32)/(x^4+16)

\int \frac{32}{x ^{4} + 16} d x

\frac{1}{2} ln x^{2} + 22 x + 4 + 2 arctan (\frac{1}{2} (x + 2)) - \frac{1}{2} ln x^{2} - 22 x + 4 + 2 arctan (\frac{1}{2} (x - 2)) + C

Solution steps

\int \frac{32}{x ^{4} + 16} d x

Take the constant out:

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

= 32 \cdot \int \frac{1}{x ^{4} + 16} d x

Take the partial fraction of

\frac{1}{x ^{4} + 16} : \frac{x + 2 2}{16 2 ( x ^{2} + 2 2 x + 4 )} + \frac{- x + 2 2}{16 2 ( x ^{2} - 2 2 x + 4 )}

= 32 \cdot \int \frac{x + 2 2}{16 2 ( x ^{2} + 2 2 x + 4 )} + \frac{- x + 2 2}{16 2 ( x ^{2} - 2 2 x + 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

= 32 (\int \frac{x + 2 2}{16 2 ( x ^{2} + 2 2 x + 4 )} d x + \int \frac{- x + 2 2}{16 2 ( x ^{2} - 2 2 x + 4 )} d x)

\int \frac{x + 2 2}{16 2 ( x ^{2} + 2 2 x + 4 )} d x = \frac{1}{32 2} (ln x^{2} + 22 x + 4 + 2 arctan (\frac{x + 2}{2}))

\int \frac{- x + 2 2}{16 2 ( x ^{2} - 2 2 x + 4 )} d x = \frac{1}{32 2} (- ln x^{2} - 22 x + 4 + 2 arctan (\frac{x - 2}{2}))

= 32 (\frac{1}{32 2} (ln x^{2} + 22 x + 4 + 2 arctan (\frac{x + 2}{2})) + \frac{1}{32 2} (- ln x^{2} - 22 x + 4 + 2 arctan (\frac{x - 2}{2})))

Simplify

32 (\frac{1}{32 2} (ln x^{2} + 22 x + 4 + 2 arctan (\frac{x + 2}{2})) + \frac{1}{32 2} (- ln x^{2} - 22 x + 4 + 2 arctan (\frac{x - 2}{2}))) : \frac{1}{2} ln x^{2} + 22 x + 4 + 2 arctan (\frac{1}{2} (x + 2)) - \frac{1}{2} ln x^{2} - 22 x + 4 + 2 arctan (\frac{1}{2} (x - 2))

= \frac{1}{2} ln x^{2} + 22 x + 4 + 2 arctan (\frac{1}{2} (x + 2)) - \frac{1}{2} ln x^{2} - 22 x + 4 + 2 arctan (\frac{1}{2} (x - 2))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 22 x + 4 + 2 arctan (\frac{1}{2} (x + 2)) - \frac{1}{2} ln x^{2} - 22 x + 4 + 2 arctan (\frac{1}{2} (x - 2)) + C

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