integral of t/(1+t^3)

解

\int \frac{t}{1 + t ^{3}} d t

- \frac{1}{3} ln ∣ t + 1 ∣ + \frac{1}{6} (ln 4 t^{2} - 4 t + 4 + 23 arctan (\frac{2 t - 1}{3})) + C

解答ステップ

\int \frac{t}{1 + t ^{3}} d t

以下の部分分数を得る:

\frac{t}{1 + t ^{3}} : - \frac{1}{3 ( t + 1 )} + \frac{t + 1}{3 ( t ^{2} - t + 1 )}

= \int - \frac{1}{3 ( t + 1 )} + \frac{t + 1}{3 ( t ^{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 \frac{1}{3 ( t + 1 )} d t + \int \frac{t + 1}{3 ( t ^{2} - t + 1 )} d t

\int \frac{1}{3 ( t + 1 )} d t = \frac{1}{3} ln ∣ t + 1 ∣

\int \frac{t + 1}{3 ( t ^{2} - t + 1 )} d t = \frac{1}{6} (ln 4 t^{2} - 4 t + 4 + 23 arctan (\frac{2 t - 1}{3}))

= - \frac{1}{3} ln ∣ t + 1 ∣ + \frac{1}{6} (ln 4 t^{2} - 4 t + 4 + 23 arctan (\frac{2 t - 1}{3}))

定数を解答に追加する

= - \frac{1}{3} ln ∣ t + 1 ∣ + \frac{1}{6} (ln 4 t^{2} - 4 t + 4 + 23 arctan (\frac{2 t - 1}{3})) + C