What is the integral of ((4x^2-8)^{3/2})/x ?

The integral of ((4x^2-8)^{3/2})/x is

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integral of ((4x^2-8)^{3/2})/x

\int \frac{( 4 x ^{2} - 8 ) ^{\frac{3}{2}}}{x} d x

\frac{8}{3} (x^{2} - 2)^{\frac{3}{2}} + 16 \cdot 2^{\frac{1}{2}} arcsec (\frac{x}{2 ^{\frac{1}{2}}}) - 16 \cdot 2^{\frac{1}{2}} (\frac{1}{2} (x^{2} - 2))^{\frac{1}{2}} + C

Solution steps

\int \frac{( 4 x ^{2} - 8 ) ^{\frac{3}{2}}}{x} d x

Apply Trigonometric Substitution

: \int 16 \cdot 2^{\frac{1}{2}} tan^{4} (u) d u

= \int 16 \cdot 2^{\frac{1}{2}} tan^{4} (u) d u

Take the constant out:

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

= 16 \cdot 2^{\frac{1}{2}} \cdot \int tan^{4} (u) d u

Apply Integral Reduction

= 16 \cdot 2^{\frac{1}{2}} (\frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u)

\int tan^{2} (u) d u = - u + tan (u)

= 16 \cdot 2^{\frac{1}{2}} (\frac{tan ^{3} ( u )}{3} - (- u + tan (u)))

Substitute back

u = arcsec (\frac{2}{2 \cdot 2 ^{\frac{1}{2}}} x)

= 16 \cdot 2^{\frac{1}{2}} \frac{tan ^{3} ( arcsec ( \frac{2}{2 \cdot 2 ^{\frac{1}{2}}} x ) )}{3} - (- arcsec (\frac{2}{2 \cdot 2 ^{\frac{1}{2}}} x) + tan (arcsec (\frac{2}{2 \cdot 2 ^{\frac{1}{2}}} x)))

Simplify

16 \cdot 2^{\frac{1}{2}} \frac{tan ^{3} ( arcsec ( \frac{2}{2 \cdot 2 ^{\frac{1}{2}}} x ) )}{3} - (- arcsec (\frac{2}{2 \cdot 2 ^{\frac{1}{2}}} x) + tan (arcsec (\frac{2}{2 \cdot 2 ^{\frac{1}{2}}} x))) : \frac{8}{3} (x^{2} - 2)^{\frac{3}{2}} + 16 \cdot 2^{\frac{1}{2}} arcsec (\frac{x}{2 ^{\frac{1}{2}}}) - 16 \cdot 2^{\frac{1}{2}} (\frac{1}{2} (x^{2} - 2))^{\frac{1}{2}}

= \frac{8}{3} (x^{2} - 2)^{\frac{3}{2}} + 16 \cdot 2^{\frac{1}{2}} arcsec (\frac{x}{2 ^{\frac{1}{2}}}) - 16 \cdot 2^{\frac{1}{2}} (\frac{1}{2} (x^{2} - 2))^{\frac{1}{2}}

Add a constant to the solution

= \frac{8}{3} (x^{2} - 2)^{\frac{3}{2}} + 16 \cdot 2^{\frac{1}{2}} arcsec (\frac{x}{2 ^{\frac{1}{2}}}) - 16 \cdot 2^{\frac{1}{2}} (\frac{1}{2} (x^{2} - 2))^{\frac{1}{2}} + C

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What is the integral of ((4x^2-8)^{3/2})/x ?
The integral of ((4x^2-8)^{3/2})/x is