What is the integral of (sqrt(x^2-484))/x ?

The integral of (sqrt(x^2-484))/x is -22arcsec(1/22 x)+sqrt(x^2-484)+C

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integral of (sqrt(x^2-484))/x

\int \frac{x ^{2} - 484}{x} d x

- 22 arcsec (\frac{1}{22} x) + x^{2} - 484 + C

Solution steps

\int \frac{x ^{2} - 484}{x} d x

Apply Trigonometric Substitution

: \int 22 tan^{2} (u) d u

= \int 22 tan^{2} (u) d u

Take the constant out:

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

= 22 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 22 \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

= 22 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 22 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{22} x)

= 22 (- arcsec (\frac{1}{22} x) + tan (arcsec (\frac{1}{22} x)))

Simplify

22 (- arcsec (\frac{1}{22} x) + tan (arcsec (\frac{1}{22} x))) : - 22 arcsec (\frac{1}{22} x) + x^{2} - 484

= - 22 arcsec (\frac{1}{22} x) + x^{2} - 484

Add a constant to the solution

= - 22 arcsec (\frac{1}{22} x) + x^{2} - 484 + C

\int_{1}^{3} (x^{2} + 2 x - 3) d x

\int x^{3} arcsin (x) d x

2.5 x^{2} + 0.36 + 4 (1.1 - x)

\int \frac{x + 4}{x ^{2} + 9 x - 10} d x

f (x) = - \frac{2}{x ^{3}}

What is the integral of (sqrt(x^2-484))/x ?
The integral of (sqrt(x^2-484))/x is -22arcsec(1/22 x)+sqrt(x^2-484)+C