What is the integral of (sqrt(x^2-36))/x ?

The integral of (sqrt(x^2-36))/x is -6arcsec(1/6 x)+sqrt(x^2-36)+C

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integral of (sqrt(x^2-36))/x

\int \frac{x ^{2} - 36}{x} d x

- 6 arcsec (\frac{1}{6} x) + x^{2} - 36 + C

Solution steps

\int \frac{x ^{2} - 36}{x} d x

Apply Trigonometric Substitution

: \int 6 tan^{2} (u) d u

= \int 6 tan^{2} (u) d u

Take the constant out:

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

= 6 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 6 \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

= 6 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 6 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{6} x)

= 6 (- arcsec (\frac{1}{6} x) + tan (arcsec (\frac{1}{6} x)))

Simplify

6 (- arcsec (\frac{1}{6} x) + tan (arcsec (\frac{1}{6} x))) : - 6 arcsec (\frac{1}{6} x) + x^{2} - 36

= - 6 arcsec (\frac{1}{6} x) + x^{2} - 36

Add a constant to the solution

= - 6 arcsec (\frac{1}{6} x) + x^{2} - 36 + C

\frac{( 1 - 3 x )}{2 + x}

\frac{d}{d x} (8 sin (x) - x)

\frac{d}{d x} (\frac{f ( x ) g ( x )}{f ( x ) - g ( x )})

\int e^{- x y} sin (x) d x

\int x sin (3 x + 1) d x

What is the integral of (sqrt(x^2-36))/x ?
The integral of (sqrt(x^2-36))/x is -6arcsec(1/6 x)+sqrt(x^2-36)+C