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Popular >

integral of 1/(x^4+16)

Solution

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

Solution

\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})) + C

Solution steps

\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 )}

= \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

= \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}))

= \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}))

Add a constant to the solution

= \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})) + C

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Frequently Asked Questions (FAQ)

What is the integral of 1/(x^4+16) ?
The integral of 1/(x^4+16) is