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

解

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

\frac{1}{2} (2 (\frac{1}{4} ln 2 x^{2} + 22 x + 2 - \frac{1}{4} ln (42 + 10)) + \frac{1}{2} (arctan (2 x + 1) - arctan (22 + 1))) + \frac{1}{2} (- 2 (\frac{1}{4} ln 2 x^{2} - 22 x + 2 - \frac{1}{4} ln (10 - 42)) + \frac{1}{2} (arctan (2 x - 1) - arctan (22 - 1)))

解答ステップ

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

以下の部分分数を得る:

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

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

\int_{2}^{x} \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} d x = \frac{1}{2} (2 (\frac{1}{4} ln 2 x^{2} + 22 x + 2 - \frac{1}{4} ln (42 + 10)) + \frac{1}{2} (arctan (2 x + 1) - arctan (22 + 1)))

\int_{2}^{x} \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )} d x = \frac{1}{2} (- 2 (\frac{1}{4} ln 2 x^{2} - 22 x + 2 - \frac{1}{4} ln (10 - 42)) + \frac{1}{2} (arctan (2 x - 1) - arctan (22 - 1)))

= \frac{1}{2} (2 (\frac{1}{4} ln 2 x^{2} + 22 x + 2 - \frac{1}{4} ln (42 + 10)) + \frac{1}{2} (arctan (2 x + 1) - arctan (22 + 1))) + \frac{1}{2} (- 2 (\frac{1}{4} ln 2 x^{2} - 22 x + 2 - \frac{1}{4} ln (10 - 42)) + \frac{1}{2} (arctan (2 x - 1) - arctan (22 - 1)))