What is the integral of x^3sqrt(x^2-1) ?

The integral of x^3sqrt(x^2-1) is 1/15 (3x^2+2)(x^2-1)^{3/2}+C

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integral of x^3sqrt(x^2-1)

\int x^{3} x^{2} - 1 d x

\frac{1}{15} (3 x^{2} + 2) (x^{2} - 1)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} - 1 d x

Apply u-substitution

: \int \frac{( u + 1 ) u}{2} d u

= \int \frac{( u + 1 ) u}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int (u + 1) u d u

Expand

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

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

Substitute back

u = x^{2} - 1

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

Simplify

\frac{1}{2} (\frac{2}{5} (x^{2} - 1)^{\frac{5}{2}} + \frac{2}{3} (x^{2} - 1)^{\frac{3}{2}}) : \frac{1}{15} (3 x^{2} + 2) (x^{2} - 1)^{\frac{3}{2}}

= \frac{1}{15} (3 x^{2} + 2) (x^{2} - 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{15} (3 x^{2} + 2) (x^{2} - 1)^{\frac{3}{2}} + C

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

\int sin^{3} (7 x) cos^{- 2} (7 x) d x

t y^{^{'}} = 2 y + 3 t^{3} cos (2 t), y (\frac{π}{4}) = 0

\frac{6 x ^{2} + 4 x + 6}{x}

\frac{\partial}{\partial y} (\frac{( x + y )}{z})

What is the integral of x^3sqrt(x^2-1) ?
The integral of x^3sqrt(x^2-1) is 1/15 (3x^2+2)(x^2-1)^{3/2}+C