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

解答

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

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

求解步骤

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

将

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

用部份分式展开:

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

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

使用积分加法定则:

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

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

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

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

解答补常数

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

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