What is the integral of (2021)/(y^4+1) ?

The integral of (2021)/(y^4+1) is

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integral of (2021)/(y^4+1)

\int \frac{2021}{y ^{4} + 1} d y

2021 (\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{2021}{y ^{4} + 1} d y

Take the constant out:

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

= 2021 \cdot \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 )}

= 2021 \cdot \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

= 2021 (\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))

= 2021 (\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

= 2021 (\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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