What is the integral of (sqrt(x^2-324))/x ?

The integral of (sqrt(x^2-324))/x is -18arcsec(1/18 x)+sqrt(x^2-324)+C

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integral of (sqrt(x^2-324))/x

\int \frac{x ^{2} - 324}{x} d x

- 18 arcsec (\frac{1}{18} x) + x^{2} - 324 + C

Solution steps

\int \frac{x ^{2} - 324}{x} d x

Apply Trigonometric Substitution

: \int 18 tan^{2} (u) d u

= \int 18 tan^{2} (u) d u

Take the constant out:

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

= 18 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 18 \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

= 18 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 18 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{18} x)

= 18 (- arcsec (\frac{1}{18} x) + tan (arcsec (\frac{1}{18} x)))

Simplify

18 (- arcsec (\frac{1}{18} x) + tan (arcsec (\frac{1}{18} x))) : - 18 arcsec (\frac{1}{18} x) + x^{2} - 324

= - 18 arcsec (\frac{1}{18} x) + x^{2} - 324

Add a constant to the solution

= - 18 arcsec (\frac{1}{18} x) + x^{2} - 324 + C

\frac{d}{d x} (3 cos (x) + 4 x^{2})

x^{4} + 5 y^{4} x^{5} + 4 y^{^{'}} = 0, y (0) = 1

\int \frac{4 x ^{4} - 5 x ^{2}}{( x + 3 ) ( x - 2 )} d x

\int x sin (n) x d x

n \to 2 lim (\frac{n - 1}{n ^{2} - 1})

What is the integral of (sqrt(x^2-324))/x ?
The integral of (sqrt(x^2-324))/x is -18arcsec(1/18 x)+sqrt(x^2-324)+C