What is the integral of 1/(x^2(sqrt(x^2-9))) ?

The integral of 1/(x^2(sqrt(x^2-9))) is (sqrt(x^2-9))/(9x)+C

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integral of 1/(x^2(sqrt(x^2-9)))

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

\frac{x ^{2} - 9}{9 x} + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{9 sec ( u )} d u

= \int \frac{1}{9 sec ( u )} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{9} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{9} sin (u)

Substitute back

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

= \frac{1}{9} sin (arcsec (\frac{1}{3} x))

Simplify

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

= \frac{x ^{2} - 9}{9 x}

Add a constant to the solution

= \frac{x ^{2} - 9}{9 x} + C

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

(1, 2), (- 1, - 2)

y = x + x, (1, 2)

\frac{d}{d x} (\frac{8 x ^{2} + 6 x + 4}{x})

\frac{d ^{2}}{d x ^{2}} (5 sec (x))

What is the integral of 1/(x^2(sqrt(x^2-9))) ?
The integral of 1/(x^2(sqrt(x^2-9))) is (sqrt(x^2-9))/(9x)+C