What is the integral of (sqrt(x^2-25))/x ?

The integral of (sqrt(x^2-25))/x is -5arcsec(1/5 x)+sqrt(x^2-25)+C

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integral of (sqrt(x^2-25))/x

\int \frac{x ^{2} - 25}{x} d x

- 5 arcsec (\frac{1}{5} x) + x^{2} - 25 + C

Solution steps

\int \frac{x ^{2} - 25}{x} d x

Apply Trigonometric Substitution

: \int 5 tan^{2} (u) d u

= \int 5 tan^{2} (u) d u

Take the constant out:

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

= 5 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 5 \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

= 5 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 5 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{5} x)

= 5 (- arcsec (\frac{1}{5} x) + tan (arcsec (\frac{1}{5} x)))

Simplify

5 (- arcsec (\frac{1}{5} x) + tan (arcsec (\frac{1}{5} x))) : - 5 arcsec (\frac{1}{5} x) + x^{2} - 25

= - 5 arcsec (\frac{1}{5} x) + x^{2} - 25

Add a constant to the solution

= - 5 arcsec (\frac{1}{5} x) + x^{2} - 25 + C

\int 1064 - 0.005 x d x

x^{2} + 4 x

\frac{8}{( 2 s + 1 ) ^{3}}

\frac{\partial}{\partial x} (2 x + 3 y - 4)

f (x) \frac{2}{x}

What is the integral of (sqrt(x^2-25))/x ?
The integral of (sqrt(x^2-25))/x is -5arcsec(1/5 x)+sqrt(x^2-25)+C