integral of sqrt(1+t^3)

解

\int 1 + t^{3} d t

\frac{2 ( t ^{4} + 6 - 1 \cdot 3 ^{\frac{3}{4}} F ( 3 - 1 ∣ arcsin ( \frac{1}{4 3} - ( - 1 ) ^{\frac{5}{6}} ( t + 1 ) ) ) - 6 - 1 ( t + 1 ) t ^{2} - t + 1 + t )}{5 t ^{3} + 1} + C

解答ステップ

\int 1 + t^{3} d t

非初等積分を使用する:

\int 1 + x^{3} d x = \frac{2 ( x ^{4} + 6 - 1 3 ^{\frac{3}{4}} - 6 - 1 ( x + ( - 1 ) ^{\frac{2}{3}} ) ( - 1 ) ^{\frac{2}{3}} x ^{2} + 3 - 1 x + 1 F ( arcsin ( \frac{- ( - 1 ) ^{\frac{5}{6}} ( x + 1 )}{4 3} ) ∣ 3 - 1 ) + x )}{5 x ^{3} + 1}

= \frac{2 ( t ^{4} + 6 - 1 \cdot 3 ^{\frac{3}{4}} - 6 - 1 ( t + ( - 1 ) ^{\frac{2}{3}} ) ( - 1 ) ^{\frac{2}{3}} t ^{2} + 3 - 1 t + 1 F ( arcsin ( \frac{- ( - 1 ) ^{\frac{5}{6}} ( t + 1 )}{4 3} ) ∣ 3 - 1 ) + t )}{5 t ^{3} + 1}

簡素化

\frac{2 ( t ^{4} + 6 - 1 \cdot 3 ^{\frac{3}{4}} - 6 - 1 ( t + ( - 1 ) ^{\frac{2}{3}} ) ( - 1 ) ^{\frac{2}{3}} t ^{2} + 3 - 1 t + 1 F ( arcsin ( \frac{- ( - 1 ) ^{\frac{5}{6}} ( t + 1 )}{4 3} ) ∣ 3 - 1 ) + t )}{5 t ^{3} + 1} : \frac{2 ( t ^{4} + 6 - 1 \cdot 3 ^{\frac{3}{4}} F ( 3 - 1 ∣ arcsin ( \frac{1}{4 3} - ( - 1 ) ^{\frac{5}{6}} ( t + 1 ) ) ) - 6 - 1 ( t + 1 ) t ^{2} - t + 1 + t )}{5 t ^{3} + 1}

= \frac{2 ( t ^{4} + 6 - 1 \cdot 3 ^{\frac{3}{4}} F ( 3 - 1 ∣ arcsin ( \frac{1}{4 3} - ( - 1 ) ^{\frac{5}{6}} ( t + 1 ) ) ) - 6 - 1 ( t + 1 ) t ^{2} - t + 1 + t )}{5 t ^{3} + 1}

定数を解答に追加する

= \frac{2 ( t ^{4} + 6 - 1 \cdot 3 ^{\frac{3}{4}} F ( 3 - 1 ∣ arcsin ( \frac{1}{4 3} - ( - 1 ) ^{\frac{5}{6}} ( t + 1 ) ) ) - 6 - 1 ( t + 1 ) t ^{2} - t + 1 + t )}{5 t ^{3} + 1} + C