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Popular >

integral of 1/(1+t^4)

Solution

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

Solution

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

Solution steps

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

Take the partial fraction of

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

= \int \frac{t + 2}{2 2 ( t ^{2} + 2 t + 1 )} + \frac{- t + 2}{2 2 ( t ^{2} - 2 t + 1 )} 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 2 ( t ^{2} + 2 t + 1 )} d t + \int \frac{- t + 2}{2 2 ( t ^{2} - 2 t + 1 )} d t

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

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

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

Add a constant to the solution

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

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Frequently Asked Questions (FAQ)

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