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Popular >

integral of 1/(y^4+1)

Solution

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

Solution

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

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Frequently Asked Questions (FAQ)

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