What is the integral of (3y^2+6y)(y^3+3y^2+1)^{2/3} ?

The integral of (3y^2+6y)(y^3+3y^2+1)^{2/3} is 3/5 (y^3+3y^2+1)^{5/3}+C

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integral of (3y^2+6y)(y^3+3y^2+1)^{2/3}

\int (3 y^{2} + 6 y) (y^{3} + 3 y^{2} + 1)^{\frac{2}{3}} d y

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

Solution steps

\int (3 y^{2} + 6 y) (y^{3} + 3 y^{2} + 1)^{\frac{2}{3}} d y

Apply u-substitution

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

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

Apply the Power Rule

: \frac{3}{5} u^{\frac{5}{3}}

= \frac{3}{5} u^{\frac{5}{3}}

Substitute back

u = y^{3} + 3 y^{2} + 1

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

Add a constant to the solution

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

f (x) = x^{2} - 8

\frac{d}{d x} (\frac{x + 3}{∣ x + 3 ∣})

x \to 0 + lim (\frac{( ln ( x ) ) ^{2}}{x})

\frac{\partial}{\partial x} (sin^{4} (x))

(- 4, 6), (3, - 8)

What is the integral of (3y^2+6y)(y^3+3y^2+1)^{2/3} ?
The integral of (3y^2+6y)(y^3+3y^2+1)^{2/3} is 3/5 (y^3+3y^2+1)^{5/3}+C