What is the integral of (sqrt(x^2-169))/x ?

The integral of (sqrt(x^2-169))/x is -13arcsec(1/13 x)+sqrt(x^2-169)+C

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integral of (sqrt(x^2-169))/x

\int \frac{x ^{2} - 169}{x} d x

- 13 arcsec (\frac{1}{13} x) + x^{2} - 169 + C

Solution steps

\int \frac{x ^{2} - 169}{x} d x

Apply Trigonometric Substitution

: \int 13 tan^{2} (u) d u

= \int 13 tan^{2} (u) d u

Take the constant out:

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

= 13 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 13 \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

= 13 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 13 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{13} x)

= 13 (- arcsec (\frac{1}{13} x) + tan (arcsec (\frac{1}{13} x)))

Simplify

13 (- arcsec (\frac{1}{13} x) + tan (arcsec (\frac{1}{13} x))) : - 13 arcsec (\frac{1}{13} x) + x^{2} - 169

= - 13 arcsec (\frac{1}{13} x) + x^{2} - 169

Add a constant to the solution

= - 13 arcsec (\frac{1}{13} x) + x^{2} - 169 + C

\frac{d}{d x} (\frac{- 2}{x ^{2}})

ln (1 + 5 x)

\frac{d y}{d x}, y = 1 3^{x}

\frac{d}{d x} (- \frac{2}{e ^{x}})

\frac{d}{d x} (\frac{cos ( x )}{e ^{x}})

What is the integral of (sqrt(x^2-169))/x ?
The integral of (sqrt(x^2-169))/x is -13arcsec(1/13 x)+sqrt(x^2-169)+C