integral from 1 to x of 5/(t^3+1)

解

\int_{1}^{x} \frac{5}{t ^{3} + 1} d t

5 (\frac{1}{3} (ln ∣ x + 1 ∣ - ln (2)) + \frac{1}{3} (- \frac{1}{2} ln (\frac{4 x ^{2} - 4 x + 4}{4}) + 3 (arctan (\frac{2 x - 1}{3}) - \frac{π}{6})))

解答ステップ

\int_{1}^{x} \frac{5}{t ^{3} + 1} d t

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int_{1}^{x} \frac{1}{t ^{3} + 1} d t

以下の部分分数を得る:

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

= 5 \cdot \int_{1}^{x} \frac{1}{3 ( t + 1 )} + \frac{- t + 2}{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

= 5 (\int_{1}^{x} \frac{1}{3 ( t + 1 )} d t + \int_{1}^{x} \frac{- t + 2}{3 ( t ^{2} - t + 1 )} d t)

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

\int_{1}^{x} \frac{- t + 2}{3 ( t ^{2} - t + 1 )} d t = \frac{1}{3} (- \frac{1}{2} ln (\frac{4 x ^{2} - 4 x + 4}{4}) + 3 (arctan (\frac{2 x - 1}{3}) - \frac{π}{6}))

= 5 (\frac{1}{3} (ln ∣ x + 1 ∣ - ln (2)) + \frac{1}{3} (- \frac{1}{2} ln (\frac{4 x ^{2} - 4 x + 4}{4}) + 3 (arctan (\frac{2 x - 1}{3}) - \frac{π}{6})))