What is the integral of 1/(x(x^2-9)^{3/2)} ?

The integral of 1/(x(x^2-9)^{3/2)} is 1/27 (-arcsec(1/3 x)-3/((x^2-9)^{1/2)})+C

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integral of 1/(x(x^2-9)^{3/2)}

\int \frac{1}{x ( x ^{2} - 9 ) ^{\frac{3}{2}}} d x

\frac{1}{27} (- arcsec (\frac{1}{3} x) - \frac{3}{( x ^{2} - 9 ) ^{\frac{1}{2}}}) + C

Solution steps

\int \frac{1}{x ( x ^{2} - 9 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{27 tan ^{2} ( u )} d u

= \int \frac{1}{27 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}{27} \cdot \int \frac{1}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{27} \cdot \int - 1 + csc^{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}{27} (- \int 1 d u + \int csc^{2} (u) d u)

\int 1 d u = u

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{27} (- u - cot (u))

Substitute back

u = arcsec (\frac{1}{3} x)

= \frac{1}{27} (- arcsec (\frac{1}{3} x) - cot (arcsec (\frac{1}{3} x)))

Simplify

\frac{1}{27} (- arcsec (\frac{1}{3} x) - cot (arcsec (\frac{1}{3} x))) : \frac{1}{27} (- arcsec (\frac{1}{3} x) - \frac{3}{( x ^{2} - 9 ) ^{\frac{1}{2}}})

= \frac{1}{27} (- arcsec (\frac{1}{3} x) - \frac{3}{( x ^{2} - 9 ) ^{\frac{1}{2}}})

Add a constant to the solution

= \frac{1}{27} (- arcsec (\frac{1}{3} x) - \frac{3}{( x ^{2} - 9 ) ^{\frac{1}{2}}}) + C

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

\frac{d y}{d x} + e^{x} y = x^{2} y^{2}

x \to - 2 lim (3 x^{3} - 4 x + 2)

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

\frac{d}{d x} (x^{3} (5.4)^{x})

What is the integral of 1/(x(x^2-9)^{3/2)} ?
The integral of 1/(x(x^2-9)^{3/2)} is 1/27 (-arcsec(1/3 x)-3/((x^2-9)^{1/2)})+C