What is the integral of x^5sqrt(4x^2-1) ?

The integral of x^5sqrt(4x^2-1) is 1/840 (30x^4+6x^2+1)(4x^2-1)^{3/2}+C

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integral of x^5sqrt(4x^2-1)

\int x^{5} 4 x^{2} - 1 d x

\frac{1}{840} (30 x^{4} + 6 x^{2} + 1) (4 x^{2} - 1)^{\frac{3}{2}} + C

Solution steps

\int x^{5} 4 x^{2} - 1 d x

Apply Trigonometric Substitution

: \int \frac{1}{64} sec^{6} (u) tan^{2} (u) d u

= \int \frac{1}{64} sec^{6} (u) tan^{2} (u) d u

Take the constant out:

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

= \frac{1}{64} \cdot \int sec^{6} (u) tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{64} \cdot \int sec^{2} (u) (1 + tan^{2} (u))^{2} tan^{2} (u) d u

Apply u-substitution

: \int v^{2} (1 + v^{2})^{2} d v

= \frac{1}{64} \cdot \int v^{2} (1 + v^{2})^{2} d v

Expand

v^{2} (1 + v^{2})^{2} : v^{2} + 2 v^{4} + v^{6}

= \frac{1}{64} \cdot \int v^{2} + 2 v^{4} + v^{6} 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}{64} (\int v^{2} d v + \int 2 v^{4} d v + \int v^{6} d v)

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

\int 2 v^{4} d v = \frac{2 v ^{5}}{5}

\int v^{6} d v = \frac{v ^{7}}{7}

= \frac{1}{64} (\frac{v ^{3}}{3} + \frac{2 v ^{5}}{5} + \frac{v ^{7}}{7})

Substitute back

= \frac{1}{64} (\frac{tan ^{3} ( arcsec ( 2 x ) )}{3} + \frac{2 tan ^{5} ( arcsec ( 2 x ) )}{5} + \frac{tan ^{7} ( arcsec ( 2 x ) )}{7})

Simplify

\frac{1}{64} (\frac{tan ^{3} ( arcsec ( 2 x ) )}{3} + \frac{2 tan ^{5} ( arcsec ( 2 x ) )}{5} + \frac{tan ^{7} ( arcsec ( 2 x ) )}{7}) : \frac{1}{840} (30 x^{4} + 6 x^{2} + 1) (4 x^{2} - 1)^{\frac{3}{2}}

= \frac{1}{840} (30 x^{4} + 6 x^{2} + 1) (4 x^{2} - 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{840} (30 x^{4} + 6 x^{2} + 1) (4 x^{2} - 1)^{\frac{3}{2}} + C

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What is the integral of x^5sqrt(4x^2-1) ?
The integral of x^5sqrt(4x^2-1) is 1/840 (30x^4+6x^2+1)(4x^2-1)^{3/2}+C