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

The integral of 1/(x^3sqrt(9x^2-1)) is 9/2 (arcsec(3x)+1/2 sin(2arcsec(3x)))+C

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

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

\frac{9}{2} (arcsec (3 x) + \frac{1}{2} sin (2 arcsec (3 x))) + C

Solution steps

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

Apply Trigonometric Substitution

: \int 9 cos^{2} (u) d u

= \int 9 cos^{2} (u) d u

Take the constant out:

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

= 9 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 9 \cdot \int \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= 9 \cdot \frac{1}{2} \cdot \int 1 + cos (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

= 9 \cdot \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= 9 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsec (3 x)

= 9 \cdot \frac{1}{2} (arcsec (3 x) + \frac{1}{2} sin (2 arcsec (3 x)))

Simplify

9 \cdot \frac{1}{2} (arcsec (3 x) + \frac{1}{2} sin (2 arcsec (3 x))) : \frac{9}{2} (arcsec (3 x) + \frac{1}{2} sin (2 arcsec (3 x)))

= \frac{9}{2} (arcsec (3 x) + \frac{1}{2} sin (2 arcsec (3 x)))

Add a constant to the solution

= \frac{9}{2} (arcsec (3 x) + \frac{1}{2} sin (2 arcsec (3 x))) + C

\frac{d}{d x} (\frac{x ^{2}}{30})

\frac{d}{d x} (x y + x cos (x))

\int x^{2} + x d x

2 y^{^{''}} + 3 y^{^{'}} - 4 y = 5 e^{3 x}

n = 1 \sum \infty \frac{( 3 x - 2 ) ^{n}}{n}

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