What is the integral of (sqrt(9x^2-4))/x ?

The integral of (sqrt(9x^2-4))/x is -2arcsec(3/2 x)+sqrt(9x^2-4)+C

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integral of (sqrt(9x^2-4))/x

\int \frac{9 x ^{2} - 4}{x} d x

- 2 arcsec (\frac{3}{2} x) + 9 x^{2} - 4 + C

Solution steps

\int \frac{9 x ^{2} - 4}{x} d x

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= 2 \cdot \int - 1 + sec^{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

= 2 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 2 (- u + tan (u))

Substitute back

u = arcsec (\frac{3}{2} x)

= 2 (- arcsec (\frac{3}{2} x) + tan (arcsec (\frac{3}{2} x)))

Simplify

2 (- arcsec (\frac{3}{2} x) + tan (arcsec (\frac{3}{2} x))) : - 2 arcsec (\frac{3}{2} x) + 9 x^{2} - 4

= - 2 arcsec (\frac{3}{2} x) + 9 x^{2} - 4

Add a constant to the solution

= - 2 arcsec (\frac{3}{2} x) + 9 x^{2} - 4 + C

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

y^{^{'}} + 3 y x^{5} = 0

\frac{d}{d x} (w (\frac{x}{50} - F (w, x)))

e^{x}, x^{2} - 1, - 1, 1

n = 0 \sum \infty \frac{6}{4 ^{n}}

What is the integral of (sqrt(9x^2-4))/x ?
The integral of (sqrt(9x^2-4))/x is -2arcsec(3/2 x)+sqrt(9x^2-4)+C