What is the integral of x/((x^2+1)(x^4+1)) ?

The integral of x/((x^2+1)(x^4+1)) is

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integral of x/((x^2+1)(x^4+1))

\int \frac{x}{( x ^{2} + 1 ) ( x ^{4} + 1 )} d x

\frac{1}{4} ln x^{2} + 1 - \frac{1}{8} (ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1)) + \frac{1}{8} (- ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)) + C

Solution steps

\int \frac{x}{( x ^{2} + 1 ) ( x ^{4} + 1 )} d x

Take the partial fraction of

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

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

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

\int \frac{x + 2}{4 ( x ^{2} + 2 x + 1 )} d x = \frac{1}{8} (ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1))

\int \frac{- x + 2}{4 ( x ^{2} - 2 x + 1 )} d x = \frac{1}{8} (- ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1))

= \frac{1}{4} ln x^{2} + 1 - \frac{1}{8} (ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1)) + \frac{1}{8} (- ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1))

Add a constant to the solution

= \frac{1}{4} ln x^{2} + 1 - \frac{1}{8} (ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1)) + \frac{1}{8} (- ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)) + C

\int \frac{5}{2} x e^{\frac{3}{2} x} d x

\int 5 x^{2} + 2 d x

\int \frac{1}{x \cdot x + 1} d x

\int 9 z d z

What is the integral of x/((x^2+1)(x^4+1)) ?
The integral of x/((x^2+1)(x^4+1)) is