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

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

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

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

2 (- \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{2 x ^{2}}{1 + x ^{4}} d x

Take the constant out:

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

= 2 \cdot \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 )}

= 2 \cdot \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

= 2 (- \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))

= 2 (- \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

= 2 (- \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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