What is the integral of y^5sqrt(y^3-2) ?

The integral of y^5sqrt(y^3-2) is 1/3 (2/5 (y^3-2)^{5/2}+4/3 (y^3-2)^{3/2})+C

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integral of y^5sqrt(y^3-2)

\int y^{5} y^{3} - 2 d y

\frac{1}{3} (\frac{2}{5} (y^{3} - 2)^{\frac{5}{2}} + \frac{4}{3} (y^{3} - 2)^{\frac{3}{2}}) + C

Solution steps

\int y^{5} y^{3} - 2 d y

Apply u-substitution

: \int \frac{( u + 2 ) u}{3} d u

= \int \frac{( u + 2 ) u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int (u + 2) u d u

Expand

(u + 2) u : u^{\frac{3}{2}} + 2 u

= \frac{1}{3} \cdot \int u^{\frac{3}{2}} + 2 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int u^{\frac{3}{2}} d u + \int 2 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 2 u d u = \frac{4}{3} u^{\frac{3}{2}}

= \frac{1}{3} (\frac{2}{5} u^{\frac{5}{2}} + \frac{4}{3} u^{\frac{3}{2}})

Substitute back

u = y^{3} - 2

= \frac{1}{3} (\frac{2}{5} (y^{3} - 2)^{\frac{5}{2}} + \frac{4}{3} (y^{3} - 2)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{3} (\frac{2}{5} (y^{3} - 2)^{\frac{5}{2}} + \frac{4}{3} (y^{3} - 2)^{\frac{3}{2}}) + C

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What is the integral of y^5sqrt(y^3-2) ?
The integral of y^5sqrt(y^3-2) is 1/3 (2/5 (y^3-2)^{5/2}+4/3 (y^3-2)^{3/2})+C