What is the integral of (x^4+1)^{-1} ?

The integral of (x^4+1)^{-1} is

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integral of (x^4+1)^{-1}

\int (x^{4} + 1)^{- 1} d x

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

Solution steps

\int (x^{4} + 1)^{- 1} d x

Simplify

x^{4} : x^{3} x

= \int (x^{3} x + 1)^{- 1} d x

Simplify

(x^{3} x + 1)^{- 1} : \frac{1}{x ^{4} + 1}

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

\int 4 x^{3} - 2 x^{2} + 4 x d x

\int (3 x + 1 - \frac{x}{x ^{2} + 3}) d x

\int \frac{1}{x} + e^{x} d x

\int (z + 1) e^{8 z} d z

\frac{1}{( x + y ) ^{2}}