What is the integral of (sqrt(x^2-121))/x ?

The integral of (sqrt(x^2-121))/x is -11arcsec(1/11 x)+sqrt(x^2-121)+C

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integral of (sqrt(x^2-121))/x

\int \frac{x ^{2} - 121}{x} d x

- 11 arcsec (\frac{1}{11} x) + x^{2} - 121 + C

Solution steps

\int \frac{x ^{2} - 121}{x} d x

Apply Trigonometric Substitution

: \int 11 tan^{2} (u) d u

= \int 11 tan^{2} (u) d u

Take the constant out:

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

= 11 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 11 \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

= 11 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 11 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{11} x)

= 11 (- arcsec (\frac{1}{11} x) + tan (arcsec (\frac{1}{11} x)))

Simplify

11 (- arcsec (\frac{1}{11} x) + tan (arcsec (\frac{1}{11} x))) : - 11 arcsec (\frac{1}{11} x) + x^{2} - 121

= - 11 arcsec (\frac{1}{11} x) + x^{2} - 121

Add a constant to the solution

= - 11 arcsec (\frac{1}{11} x) + x^{2} - 121 + C

k = 1 \sum \infty \frac{( - 1 ) ^{k} \cdot ( 1 ) ^{(3 \cdot k)}}{k}

\int x y (x + y) d y

n = 0 \sum \infty \frac{2 ^{n}}{n !}

x (x - 2) y^{^{'}} + 2 y = 0

\int (x + 2) x - 2 d x

What is the integral of (sqrt(x^2-121))/x ?
The integral of (sqrt(x^2-121))/x is -11arcsec(1/11 x)+sqrt(x^2-121)+C