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

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

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

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

Take the partial fraction of

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

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

\int \frac{x}{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 ( 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

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