What is the integral of (sqrt(x^2-361))/x ?

The integral of (sqrt(x^2-361))/x is -19arcsec(1/19 x)+sqrt(x^2-361)+C

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integral of (sqrt(x^2-361))/x

\int \frac{x ^{2} - 361}{x} d x

- 19 arcsec (\frac{1}{19} x) + x^{2} - 361 + C

Solution steps

\int \frac{x ^{2} - 361}{x} d x

Apply Trigonometric Substitution

: \int 19 tan^{2} (u) d u

= \int 19 tan^{2} (u) d u

Take the constant out:

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

= 19 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 19 \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

= 19 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 19 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{19} x)

= 19 (- arcsec (\frac{1}{19} x) + tan (arcsec (\frac{1}{19} x)))

Simplify

19 (- arcsec (\frac{1}{19} x) + tan (arcsec (\frac{1}{19} x))) : - 19 arcsec (\frac{1}{19} x) + x^{2} - 361

= - 19 arcsec (\frac{1}{19} x) + x^{2} - 361

Add a constant to the solution

= - 19 arcsec (\frac{1}{19} x) + x^{2} - 361 + C

\int_{0}^{2} 36 x + 145 d x

t^{2} + 6 t - 17

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

\int_{0}^{+ oo} x \cdot (x + 1)^{- 2} d x

f (x) = 5 - x

What is the integral of (sqrt(x^2-361))/x ?
The integral of (sqrt(x^2-361))/x is -19arcsec(1/19 x)+sqrt(x^2-361)+C