What is the integral of 3y^5sqrt(y^3+1) ?

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

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integral of 3y^5sqrt(y^3+1)

\int 3 y^{5} y^{3} + 1 d y

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

Solution steps

\int 3 y^{5} y^{3} + 1 d y

Take the constant out:

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

= 3 \cdot \int y^{5} y^{3} + 1 d y

Apply u-substitution

: \int \frac{( u - 1 ) u}{3} d u

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

Take the constant out:

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

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

Expand

(u - 1) u : u^{\frac{3}{2}} - u

= 3 \cdot \frac{1}{3} \cdot \int u^{\frac{3}{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

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

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

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

= 3 \cdot \frac{1}{3} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = y^{3} + 1

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

Simplify

3 \cdot \frac{1}{3} (\frac{2}{5} (y^{3} + 1)^{\frac{5}{2}} - \frac{2}{3} (y^{3} + 1)^{\frac{3}{2}}) : \frac{2}{5} (y^{3} + 1)^{\frac{5}{2}} - \frac{2}{3} (y^{3} + 1)^{\frac{3}{2}}

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

Add a constant to the solution

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

x \to - 1 lim (x + 2 x^{4})

\frac{d}{d x} (a x e^{- x})

\frac{\partial}{\partial x} (ln (a x + b y))

n = 1 \sum \infty (1)^{n}

\int (x^{2} + 1)^{2} d x

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