integral of (-x^6-x^4+x^2+6)/(x^2+1)

解答

\int \frac{- x ^{6} - x ^{4} + x ^{2} + 6}{x ^{2} + 1} d x

- \frac{x ^{5}}{5} + x + 5 arctan (x) + C

求解步骤

\int \frac{- x ^{6} - x ^{4} + x ^{2} + 6}{x ^{2} + 1} d x

乘开

\frac{- x ^{6} - x ^{4} + x ^{2} + 6}{x ^{2} + 1} : - \frac{x ^{6}}{x ^{2} + 1} - \frac{x ^{4}}{x ^{2} + 1} + \frac{x ^{2}}{x ^{2} + 1} + \frac{6}{x ^{2} + 1}

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 ^{6}}{x ^{2} + 1} d x - \int \frac{x ^{4}}{x ^{2} + 1} d x + \int \frac{x ^{2}}{x ^{2} + 1} d x + \int \frac{6}{x ^{2} + 1} d x

\int \frac{x ^{6}}{x ^{2} + 1} d x = \frac{x ^{5}}{5} - \frac{x ^{3}}{3} + x - arctan (x)

\int \frac{x ^{4}}{x ^{2} + 1} d x = \frac{x ^{3}}{3} - x + arctan (x)

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

\int \frac{6}{x ^{2} + 1} d x = 6 arctan (x)

= - (\frac{x ^{5}}{5} - \frac{x ^{3}}{3} + x - arctan (x)) - (\frac{x ^{3}}{3} - x + arctan (x)) - arctan (x) + x + 6 arctan (x)

化简

- (\frac{x ^{5}}{5} - \frac{x ^{3}}{3} + x - arctan (x)) - (\frac{x ^{3}}{3} - x + arctan (x)) - arctan (x) + x + 6 arctan (x) : - \frac{x ^{5}}{5} + x + 5 arctan (x)

= - \frac{x ^{5}}{5} + x + 5 arctan (x)

解答补常数

= - \frac{x ^{5}}{5} + x + 5 arctan (x) + C