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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Expand

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

= \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

= \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}}

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

Substitute back

u = 1 + x^{3}

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

Add a constant to the solution

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

x^{2} + 4 x - 11 = - 2 y^{2} - 3 y, (2, - 1)

\int (3 - x)^{- \frac{1}{2}} d x

\frac{d}{d x} (x^{2} + x + 4)

\frac{d h}{d t} = 4 - 16 t^{2}, h (1) = 0

\frac{d}{d x} (ln (x^{2} - 16 x))

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