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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int a^{3} tan^{4} (u) d u

= \int a^{3} tan^{4} (u) d u

Take the constant out:

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

= a^{3} \cdot \int tan^{4} (u) d u

Apply Integral Reduction

= a^{3} (\frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u)

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

= a^{3} (\frac{tan ^{3} ( u )}{3} - (- u + tan (u)))

Substitute back

u = arcsec (\frac{1}{a} x)

= a^{3} (\frac{tan ^{3} ( arcsec ( \frac{1}{a} x ) )}{3} - (- arcsec (\frac{1}{a} x) + tan (arcsec (\frac{1}{a} x))))

Simplify

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

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

Add a constant to the solution

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

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y = 5 x^{8} - 3 x^{5} + 2, (0, 4)

\frac{d}{d x} (θ (x)), tan (θ (x)) = \frac{x}{35}

(x) \frac{d y}{d x} + y = \frac{1}{y ^{2}}

\frac{x ^{2} - 9 x + 20}{( x + 2 ) ( x - 2 )}

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