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

The integral of x^5(-3+2x^2)^{3/2} is 1/630 (35x^4+30x^2+18)(2x^2-3)^{5/2}+C

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integral of x^5(-3+2x^2)^{3/2}

\int x^{5} (- 3 + 2 x^{2})^{\frac{3}{2}} d x

\frac{1}{630} (35 x^{4} + 30 x^{2} + 18) (2 x^{2} - 3)^{\frac{5}{2}} + C

Solution steps

\int x^{5} (- 3 + 2 x^{2})^{\frac{3}{2}} d x

Apply Trigonometric Substitution

: \int \frac{81 \cdot 3 ^{\frac{1}{2}}}{8} sec^{6} (u) tan^{4} (u) d u

= \int \frac{81 \cdot 3 ^{\frac{1}{2}}}{8} sec^{6} (u) tan^{4} (u) d u

Take the constant out:

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

= \frac{81 \cdot 3 ^{\frac{1}{2}}}{8} \cdot \int sec^{6} (u) tan^{4} (u) d u

Rewrite using trig identities

= \frac{81 \cdot 3 ^{\frac{1}{2}}}{8} \cdot \int sec^{2} (u) (1 + tan^{2} (u))^{2} tan^{4} (u) d u

Apply u-substitution

: \int v^{4} (1 + v^{2})^{2} d v

= \frac{81 \cdot 3 ^{\frac{1}{2}}}{8} \cdot \int v^{4} (1 + v^{2})^{2} d v

Expand

v^{4} (1 + v^{2})^{2} : v^{4} + 2 v^{6} + v^{8}

= \frac{81 \cdot 3 ^{\frac{1}{2}}}{8} \cdot \int v^{4} + 2 v^{6} + v^{8} 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{81 \cdot 3 ^{\frac{1}{2}}}{8} (\int v^{4} d v + \int 2 v^{6} d v + \int v^{8} d v)

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

\int 2 v^{6} d v = \frac{2 v ^{7}}{7}

\int v^{8} d v = \frac{v ^{9}}{9}

= \frac{81 \cdot 3 ^{\frac{1}{2}}}{8} (\frac{v ^{5}}{5} + \frac{2 v ^{7}}{7} + \frac{v ^{9}}{9})

Substitute back

= \frac{81 \cdot 3 ^{\frac{1}{2}}}{8} \frac{tan ^{5} ( arcsec ( \frac{2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}} x ) )}{5} + \frac{2 tan ^{7} ( arcsec ( \frac{2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}} x ) )}{7} + \frac{tan ^{9} ( arcsec ( \frac{2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}} x ) )}{9}

Simplify

\frac{81 \cdot 3 ^{\frac{1}{2}}}{8} \frac{tan ^{5} ( arcsec ( \frac{2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}} x ) )}{5} + \frac{2 tan ^{7} ( arcsec ( \frac{2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}} x ) )}{7} + \frac{tan ^{9} ( arcsec ( \frac{2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}} x ) )}{9} : \frac{1}{630} (35 x^{4} + 30 x^{2} + 18) (2 x^{2} - 3)^{\frac{5}{2}}

= \frac{1}{630} (35 x^{4} + 30 x^{2} + 18) (2 x^{2} - 3)^{\frac{5}{2}}

Add a constant to the solution

= \frac{1}{630} (35 x^{4} + 30 x^{2} + 18) (2 x^{2} - 3)^{\frac{5}{2}} + C

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\int (2 x + 2)^{2} d x

\int x^{2} u

What is the integral of x^5(-3+2x^2)^{3/2} ?
The integral of x^5(-3+2x^2)^{3/2} is 1/630 (35x^4+30x^2+18)(2x^2-3)^{5/2}+C