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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

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

\int_{0}^{x} \frac{t + 2}{2 2 ( t ^{2} + 2 t + 1 )} d t = \frac{1}{2} (\frac{1}{2 2} (ln 2 x^{2} + 22 x + 2 - ln (2)) + \frac{1}{2} (arctan (2 x + 1) - \frac{π}{4}))

\int_{0}^{x} \frac{- t + 2}{2 2 ( t ^{2} - 2 t + 1 )} d t = \frac{1}{2} (- \frac{1}{2 2} (ln 2 x^{2} - 22 x + 2 - ln (2)) + \frac{1}{2} (arctan (2 x - 1) + \frac{π}{4}))

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

簡素化

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

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