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

The integral of x^5sqrt(x^3+5) is 1/3 (2/5 (x^3+5)^{5/2}-10/3 (x^3+5)^{3/2})+C

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

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

\frac{1}{3} (\frac{2}{5} (x^{3} + 5)^{\frac{5}{2}} - \frac{10}{3} (x^{3} + 5)^{\frac{3}{2}}) + C

Solution steps

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

Apply u-substitution

: \int \frac{( u - 5 ) u}{3} d u

= \int \frac{( u - 5 ) 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 - 5) u d u

Expand

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

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

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

\int 5 u d u = \frac{10}{3} u^{\frac{3}{2}}

= \frac{1}{3} (\frac{2}{5} u^{\frac{5}{2}} - \frac{10}{3} u^{\frac{3}{2}})

Substitute back

u = x^{3} + 5

= \frac{1}{3} (\frac{2}{5} (x^{3} + 5)^{\frac{5}{2}} - \frac{10}{3} (x^{3} + 5)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{3} (\frac{2}{5} (x^{3} + 5)^{\frac{5}{2}} - \frac{10}{3} (x^{3} + 5)^{\frac{3}{2}}) + C

\int \frac{( x ^{3} + 64 )}{x ^{2} + 64} d x

\frac{1}{s ( s ^{2} + 34.5 s + 1000 )}

\int_{0}^{7} 1

x \to 5 lim (x^{2} - 2 x - 15)

(x (x^{2} + 4))^{\frac{1}{3}}

What is the integral of x^5sqrt(x^3+5) ?
The integral of x^5sqrt(x^3+5) is 1/3 (2/5 (x^3+5)^{5/2}-10/3 (x^3+5)^{3/2})+C