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

解答

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

\frac{- 4 arctan ( - 2 + 1 ) + 4 arctan ( 2 - 1 ) + 8 arctan ( 2 + 1 )}{4 2}

十进制

2.22144 \dots

求解步骤

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

乘开

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

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_{- 1}^{1} \frac{1}{1 + x ^{4}} d x + \int_{- 1}^{1} \frac{x ^{2}}{1 + x ^{4}} d x

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

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

= \frac{2 ln ( 3 + 2 2 ) + 2 arctan ( 2 - 1 ) + 4 arctan ( 2 + 1 ) - 2 arctan ( - 2 + 1 )}{4 2} + \frac{4 arctan ( 2 + 1 ) - 2 ln ( 3 + 2 2 ) + 2 arctan ( 2 - 1 ) - 2 arctan ( - 2 + 1 )}{4 2}

化简

\frac{2 ln ( 3 + 2 2 ) + 2 arctan ( 2 - 1 ) + 4 arctan ( 2 + 1 ) - 2 arctan ( - 2 + 1 )}{4 2} + \frac{4 arctan ( 2 + 1 ) - 2 ln ( 3 + 2 2 ) + 2 arctan ( 2 - 1 ) - 2 arctan ( - 2 + 1 )}{4 2} : \frac{- 4 arctan ( - 2 + 1 ) + 4 arctan ( 2 - 1 ) + 8 arctan ( 2 + 1 )}{4 2}

= \frac{- 4 arctan ( - 2 + 1 ) + 4 arctan ( 2 - 1 ) + 8 arctan ( 2 + 1 )}{4 2}

\int_{- 3}^{3} 12 - x^{2} - x^{2} + 6 d x

\int_{y^{2}}^{2 - y} (6 x + 2 y^{2}) d x

\int_{0}^{1} x \cdot 4 (x - x^{3}) d x

\int_{0}^{π} cos (n t) d t

\int_{1}^{5} (\frac{1}{x}) d x