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

The integral of (sqrt(x^2-36))/(2x) is 3(-arcsec(1/6 x)+1/6 sqrt(x^2-36))+C

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

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

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

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{x ^{2} - 36}{x} d x

Apply Trigonometric Substitution

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

= \frac{1}{2} \cdot \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

= \frac{1}{2} \cdot 6 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot 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

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

\int 1 d u = u

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

= \frac{1}{2} \cdot 6 (- u + tan (u))

Substitute back

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

= \frac{1}{2} \cdot 6 (- arcsec (\frac{1}{6} x) + tan (arcsec (\frac{1}{6} x)))

Simplify

\frac{1}{2} \cdot 6 (- arcsec (\frac{1}{6} x) + tan (arcsec (\frac{1}{6} x))) : 3 (- arcsec (\frac{1}{6} x) + \frac{1}{6} x^{2} - 36)

= 3 (- arcsec (\frac{1}{6} x) + \frac{1}{6} x^{2} - 36)

Add a constant to the solution

= 3 (- arcsec (\frac{1}{6} x) + \frac{1}{6} x^{2} - 36) + C

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What is the integral of (sqrt(x^2-36))/(2x) ?
The integral of (sqrt(x^2-36))/(2x) is 3(-arcsec(1/6 x)+1/6 sqrt(x^2-36))+C