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

The integral of x^5sqrt(x^3-1) is 1/45 (6(x^3-1)^{5/2}+10(x^3-1)^{3/2})+C

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

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

\frac{1}{45} (6 (x^{3} - 1)^{\frac{5}{2}} + 10 (x^{3} - 1)^{\frac{3}{2}}) + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{\frac{1}{5}} u ^{\frac{3}{5}} - 1}{5} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{5 ( v + 1 ) v}{3} d v

= \frac{1}{5} \cdot \int \frac{5 ( v + 1 ) v}{3} d v

Take the constant out:

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

= \frac{1}{5} \cdot \frac{5}{3} \cdot \int (v + 1) v d v

Expand

(v + 1) v : v^{\frac{3}{2}} + v

= \frac{1}{5} \cdot \frac{5}{3} \cdot \int v^{\frac{3}{2}} + v d v

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}{5} \cdot \frac{5}{3} (\int v^{\frac{3}{2}} d v + \int v d v)

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

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

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

Substitute back

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

Simplify

\frac{1}{5} \cdot \frac{5}{3} (\frac{2}{5} ((x^{5})^{\frac{3}{5}} - 1)^{\frac{5}{2}} + \frac{2}{3} ((x^{5})^{\frac{3}{5}} - 1)^{\frac{3}{2}}) : \frac{1}{45} (6 (x^{3} - 1)^{\frac{5}{2}} + 10 (x^{3} - 1)^{\frac{3}{2}})

= \frac{1}{45} (6 (x^{3} - 1)^{\frac{5}{2}} + 10 (x^{3} - 1)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{45} (6 (x^{3} - 1)^{\frac{5}{2}} + 10 (x^{3} - 1)^{\frac{3}{2}}) + C

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x \to - 2 lim (x^{2} - 3 x + 5)

\int_{2}^{x} 1 + t^{2} d t

y = (a x + b)^{6}

What is the integral of x^5sqrt(x^3-1) ?
The integral of x^5sqrt(x^3-1) is 1/45 (6(x^3-1)^{5/2}+10(x^3-1)^{3/2})+C