integral of sqrt(t^3+1)

Lösung

\int t^{3} + 1 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

Schritte zur Lösung

\int t^{3} + 1 d t

Das ist ein nicht elementares Integral :

\int x^{3} + 1 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}

Vereinfache

\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}

Füge eine Konstante zur Lösung hinzu

= \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

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