integral from 0 to 3 of 1/(x^4+1)

解

\int_{0}^{3} \frac{1}{x ^{4} + 1} d x

\frac{2 ln ( 3 2 + 10 ) - 2 ln ( 10 - 3 2 ) + 4 arctan ( 3 2 + 1 ) + 4 arctan ( 3 2 - 1 )}{8 2}

十進法表記

1.09843 \dots

解答ステップ

\int_{0}^{3} \frac{1}{x ^{4} + 1} d x

以下の部分分数を得る:

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

= \int_{0}^{3} \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} + \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )} d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{3} \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} d x + \int_{0}^{3} \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )} d x

\int_{0}^{3} \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} d x = \frac{1}{2} (\frac{1}{2 2} ln (32 + 10) + \frac{1}{2} (arctan (32 + 1) - \frac{π}{4}))

\int_{0}^{3} \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )} d x = \frac{1}{2} (- \frac{1}{2 2} ln (10 - 32) + \frac{1}{2} (arctan (32 - 1) + \frac{π}{4}))

= \frac{1}{2} (\frac{1}{2 2} ln (32 + 10) + \frac{1}{2} (arctan (32 + 1) - \frac{π}{4})) + \frac{1}{2} (- \frac{1}{2 2} ln (10 - 32) + \frac{1}{2} (arctan (32 - 1) + \frac{π}{4}))

簡素化

\frac{1}{2} (\frac{1}{2 2} ln (32 + 10) + \frac{1}{2} (arctan (32 + 1) - \frac{π}{4})) + \frac{1}{2} (- \frac{1}{2 2} ln (10 - 32) + \frac{1}{2} (arctan (32 - 1) + \frac{π}{4})) : \frac{2 ln ( 3 2 + 10 ) - 2 ln ( 10 - 3 2 ) + 4 arctan ( 3 2 + 1 ) + 4 arctan ( 3 2 - 1 )}{8 2}

= \frac{2 ln ( 3 2 + 10 ) - 2 ln ( 10 - 3 2 ) + 4 arctan ( 3 2 + 1 ) + 4 arctan ( 3 2 - 1 )}{8 2}