What is the integral of (sqrt(x^2-5))/x ?

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

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integral of (sqrt(x^2-5))/x

\int \frac{x ^{2} - 5}{x} d x

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

Solution steps

\int \frac{x ^{2} - 5}{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) + \frac{1}{5} (x^{2} - 5))

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (2 y^{2})

\frac{d}{d x} (2^{\frac{1}{2}})

\int tan^{6} (x) sec^{4} (x) d x

\frac{d}{d x} (4^{(x^{9})})

\int (2 x^{2} - 1)^{2} d x

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