What is the integral of y^2sqrt(2y^3+1) ?

The integral of y^2sqrt(2y^3+1) is 1/9 (2y^3+1)^{3/2}+C

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integral of y^2sqrt(2y^3+1)

\int y^{2} 2 y^{3} + 1 d y

\frac{1}{9} (2 y^{3} + 1)^{\frac{3}{2}} + C

Solution steps

\int y^{2} 2 y^{3} + 1 d y

Apply u-substitution

: \int \frac{u}{6} d u

= \int \frac{u}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 2 y^{3} + 1

= \frac{1}{6} \cdot \frac{2}{3} (2 y^{3} + 1)^{\frac{3}{2}}

Simplify

\frac{1}{6} \cdot \frac{2}{3} (2 y^{3} + 1)^{\frac{3}{2}} : \frac{1}{9} (2 y^{3} + 1)^{\frac{3}{2}}

= \frac{1}{9} (2 y^{3} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{9} (2 y^{3} + 1)^{\frac{3}{2}} + C

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

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

x \to 0 lim (\frac{1}{2 x ^{3}})

y = \frac{4 x ^{2} + 6 x}{3 x ^{3} + 2}

x \to \infty lim ((\frac{( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} )}) + 4)

What is the integral of y^2sqrt(2y^3+1) ?
The integral of y^2sqrt(2y^3+1) is 1/9 (2y^3+1)^{3/2}+C