What is the integral of (sqrt(x^2-225))/x ?

The integral of (sqrt(x^2-225))/x is -15arcsec(1/15 x)+sqrt(x^2-225)+C

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integral of (sqrt(x^2-225))/x

\int \frac{x ^{2} - 225}{x} d x

- 15 arcsec (\frac{1}{15} x) + x^{2} - 225 + C

Solution steps

\int \frac{x ^{2} - 225}{x} d x

Apply Trigonometric Substitution

: \int 15 tan^{2} (u) d u

= \int 15 tan^{2} (u) d u

Take the constant out:

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

= 15 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 15 \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

= 15 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 15 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{15} x)

= 15 (- arcsec (\frac{1}{15} x) + tan (arcsec (\frac{1}{15} x)))

Simplify

15 (- arcsec (\frac{1}{15} x) + tan (arcsec (\frac{1}{15} x))) : - 15 arcsec (\frac{1}{15} x) + x^{2} - 225

= - 15 arcsec (\frac{1}{15} x) + x^{2} - 225

Add a constant to the solution

= - 15 arcsec (\frac{1}{15} x) + x^{2} - 225 + C

y^{^{'}} = 2 x 1 - y^{2}

\frac{d}{d x} ((A x + B) e^{x})

y^{^{'}} + y = e^{9 t}, y (0) = - 1

x \to 2 lim (\frac{x + 3}{x + 2})

\frac{\partial}{\partial y} (x y^{3} + 4 y e^{x} - 2 y^{2})

What is the integral of (sqrt(x^2-225))/x ?
The integral of (sqrt(x^2-225))/x is -15arcsec(1/15 x)+sqrt(x^2-225)+C