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

解

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

\frac{1}{2} (\frac{1}{2 2} (ln 8 x^{2} + 42 x + 2 - ln (22 + 4)) + \frac{1}{2} (arctan (22 x + 1) - arctan (2 + 1))) + \frac{1}{2} (- \frac{1}{2 2} (ln 8 x^{2} - 42 x + 2 - ln (4 - 22)) + \frac{1}{2} (arctan (22 x - 1) - arctan (2 - 1)))

解答ステップ

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

以下の部分分数を得る:

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

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

\int_{1}^{2 x} \frac{t + 2}{2 2 ( t ^{2} + 2 t + 1 )} d t = \frac{1}{2} (\frac{1}{2 2} (ln 8 x^{2} + 42 x + 2 - ln (22 + 4)) + \frac{1}{2} (arctan (22 x + 1) - arctan (2 + 1)))

\int_{1}^{2 x} \frac{- t + 2}{2 2 ( t ^{2} - 2 t + 1 )} d t = \frac{1}{2} (- \frac{1}{2 2} (ln 8 x^{2} - 42 x + 2 - ln (4 - 22)) + \frac{1}{2} (arctan (22 x - 1) - arctan (2 - 1)))

= \frac{1}{2} (\frac{1}{2 2} (ln 8 x^{2} + 42 x + 2 - ln (22 + 4)) + \frac{1}{2} (arctan (22 x + 1) - arctan (2 + 1))) + \frac{1}{2} (- \frac{1}{2 2} (ln 8 x^{2} - 42 x + 2 - ln (4 - 22)) + \frac{1}{2} (arctan (22 x - 1) - arctan (2 - 1)))