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Popular >

integral of 1/(t^4+16)

Solution

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

Solution

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

Solution steps

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

Take the partial fraction of

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

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

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{t + 2 2}{16 2 ( t ^{2} + 2 2 t + 4 )} d t + \int \frac{- t + 2 2}{16 2 ( t ^{2} - 2 2 t + 4 )} d t

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

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

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

Add a constant to the solution

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

Graph

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Frequently Asked Questions (FAQ)

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