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Popular >

integral of sqrt(1+y^3)

Solution

integral

1 + y^{3}

Solution

\frac{2 ( 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 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}

= \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

\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 ( 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 ( 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 ( 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

Popular Examples

integral of (1+x+x^2)/(x^2(1+x))integral

\frac{1 + x + x ^{2}}{x ^{2} ( 1 + x )}

integral of a+x^2 integral

a + x^{2}

integral of 1/(cos^4(x))integral

\frac{1}{cos ^{4} ( x )}

integral of ((x^2-1))/((x-2))integral

\frac{( x ^{2} - 1 )}{( x - 2 )}

integral of (1+1)/x integral

\frac{1 + 1}{x}

Frequently Asked Questions (FAQ)

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