What is the integral of xsqrt(1+y^3) ?

The integral of xsqrt(1+y^3) is

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integral of xsqrt(1+y^3)

\int x 1 + y^{3} d y

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

Solution steps

\int x 1 + y^{3} d y

Take the constant out:

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

= x \cdot \int 1 + y^{3} d y

Use the nonelementary integral:

\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}

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

Simplify

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

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

Add a constant to the solution

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

\int (x^{8} - 4 x^{3} - x) d x

f (x) = \frac{8 + x}{1 - 8 x}

t \to π lim (cos (\frac{t}{3}))

\int_{0}^{\infty} e^{- 7 x} d x

f (x) = 2 x + \frac{x ^{2} - x}{x ^{3} + x + 1}