What is the integral of (sqrt(x^2-100))/x ?

The integral of (sqrt(x^2-100))/x is -10arcsec(1/10 x)+sqrt(x^2-100)+C

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integral of (sqrt(x^2-100))/x

\int \frac{x ^{2} - 100}{x} d x

- 10 arcsec (\frac{1}{10} x) + x^{2} - 100 + C

Solution steps

\int \frac{x ^{2} - 100}{x} d x

Apply Trigonometric Substitution

: \int 10 tan^{2} (u) d u

= \int 10 tan^{2} (u) d u

Take the constant out:

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

= 10 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 10 \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

= 10 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 10 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{10} x)

= 10 (- arcsec (\frac{1}{10} x) + tan (arcsec (\frac{1}{10} x)))

Simplify

10 (- arcsec (\frac{1}{10} x) + tan (arcsec (\frac{1}{10} x))) : - 10 arcsec (\frac{1}{10} x) + x^{2} - 100

= - 10 arcsec (\frac{1}{10} x) + x^{2} - 100

Add a constant to the solution

= - 10 arcsec (\frac{1}{10} x) + x^{2} - 100 + C

\frac{d y}{d x} = (y sin (x))

\int sin (12 x) cos (7 x) d x

2 t + 1

x^{2} + 6, 7 x^{2}

\frac{d}{d x} ((x^{2} + y^{2} + z^{2})^{\frac{3}{2}})

What is the integral of (sqrt(x^2-100))/x ?
The integral of (sqrt(x^2-100))/x is -10arcsec(1/10 x)+sqrt(x^2-100)+C